Namomo Spring Camp 2022 Div1 每日一题
配合食用的起手板 <a class="link" href="https://github.com/ACRei/Algorithm/blob/master/DefaultCode.cpp" target="_blank" rel="noopener"
>https://github.com/ACRei/Algorithm/blob/master/DefaultCode.cpp
Day1 P436. 子串的最大差
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const int MAXN=(int)5e5+3;
int n,ans;
int a[MAXN],fa[MAXN];
int cal(int a[],int sum=0) {
stack<pair<int,int> >s;
a[n+1]=1e8;
s.push({0,1e9});
for(int i=0;i<=n+1;i++) {
while(a[i]>s.top().se) {
auto p=s.top(); s.pop();
auto p2=s.top(); sum+=p.se*(i-p.fi)*(p.fi-p2.fi);
}
s.push({i,a[i]});
}
return sum;
}
inline void work(signed CASE=1,bool FINAL_CASE=false) {
n=read();
for(int i=1;i<=n;i++) {
a[i]=read(); fa[i]=-a[i];
}
printf("%lld\n",cal(a)+cal(fa));
return;
}
Day2 P437. no crossing
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const int MAXN=(int)103;
vector<pair<int, int> > grap[MAXN];
int n, k, m, ans=INF;
int dp[MAXN][MAXN][MAXN][MAXN];
int f(int c, int s, int e, int t){
if(~dp[c][s][e][t]) return dp[c][s][e][t];
if(!t) return 0;
int res = INF;
for(auto j : grap[c]){
if(j.se > c && j.se <= e){
res = min(min(res, f(j.se, j.se, e, t-1) + j.fi), f(j.se, c+1, j.se, t-1) + j.fi);
}
if(j.se < c && j.se >= s){
res = min(min(res, f(j.se, s, j.se, t-1) + j.fi), f(j.se, j.se, c-1, t-1) + j.fi);
}
}
return dp[c][s][e][t] = res;
}
inline void work(signed CASE=1,bool FINAL_CASE=false) {
n=read(); k=read()-1; m=read();
for(int i=1; i<=m; i++) {
int u=read(), v=read(), c=read();
grap[u].push_back(mkp(c, v));
}
memset(dp, -1, sizeof(dp));
for(int i=1; i<=n; i++) {
ans = min(min(ans, f(i, i, n, k)), f(i, 1, i, k));
}
printf("%d\n",ans>1e8?-1:ans);
return;
}
Day3 P451. Dis
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const int MAXN=(int)2e5+3;
int n,m;
int a[MAXN],pre[MAXN];
//Begin HLD 板子
int cnt,dep[MAXN],siz[MAXN],son[MAXN],top[MAXN],fa[MAXN],dfn[MAXN],rnk[MAXN];
vector<int> grap[MAXN];
int dfs1(int u) {
son[u]=-1; siz[u]=1; dep[u]=dep[fa[u]]+1;
pre[u]=pre[fa[u]]^a[u];
for(auto v:grap[u]) {
if(v==fa[u]) continue;
fa[v]=u; siz[u]+=dfs1(v);
if(son[u]==-1 || siz[son[u]]<siz[v])
son[u]=v;
}
return siz[u];
}
void dfs2(int u,int t) {
top[u]=t; dfn[u]=++cnt; rnk[cnt]=u;
if (son[u] == -1) return;
dfs2(son[u],t);
// 优先对重儿子进行 DFS，可以保证同一条重链上的点 DFS 序连续
for(int v : grap[u]) {
if(v!=fa[u] && v!=son[u])
dfs2(v,v);
}
}
int lca(int u, int v) {
while (top[u] != top[v]) {
if (dep[top[u]] > dep[top[v]]) u = fa[top[u]];
else v = fa[top[v]];
}
return dep[u] > dep[v] ? v : u;
}
//End HLD 板子
int solve(int u,int v) {
int c=lca(u,v);
int f=fa[c];
return (pre[u]^pre[f]^pre[c]^pre[v]);
}
inline void work(signed CASE=1,bool FINAL_CASE=false) {
n=read(); m=read();
for(int i=1;i<=n;i++) {
a[i]=read();
}
for(int u,v,i=1;i<n;i++) {
grap[u=read()].pb(v=read());
grap[v].pb(u);
}
dfs1(1); dfs2(1,1);
while(m--) {
printf("%d\n",solve(read(),read()));
}
return;
}
Day4 P456. 选数
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const int MAXN=(int)1e5+3;
int n;
int a[MAXN],sum[MAXN];
map<int,int> mp;
inline void work(signed CASE=1,bool FINAL_CASE=false) {
n=read();
for(int i=1;i<=n;i++) {
a[i]=(read()+a[i-1])%n;
}
mp[0]=0;
for(int i=1;i<=n;i++) {
if(mp.count(a[i])) {
printf("%d\n",i-mp[a[i]]);
for(int j=mp[a[i]]+1;j<=i;j++) {
printf("%d%c",j," \n"[j==i]);
}
return;
}
mp[a[i]]=i;
}
puts("-1");
return;
}
Day5 P452. 序列操作
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const int MAXN=(int)1e6+3;
int n,q;
int a[MAXN];
struct Node {
int lazy;
}tr[MAXN<<2];
void build(int nowl,int nowr,int rt) {
// printf("%d %d %d\n",nowl,nowr,rt);
if(nowl==nowr) {
tr[rt].lazy=a[nowl];
// printf("SET val:%d in rt:%d\n",tr[rt].lazy,rt);
return;
}
int nowm=(nowl+nowr)>>1,rx=rt<<1;
build(nowl,nowm,rx); build(nowm+1,nowr,rx|1);
}
void pushdown(int rt) {
int rx=rt<<1;
tr[rx].lazy=max(tr[rx].lazy,tr[rt].lazy);
tr[rx|1].lazy=max(tr[rx|1].lazy,tr[rt].lazy);
tr[rt].lazy=0;
}
void modify(int target,int nowl,int nowr,int rt,int val) {
if(nowl>target || nowr<target) {
return;
}
if(nowl==nowr) {
tr[rt].lazy=val;
return;
}
int nowm=(nowl+nowr)>>1,rx=rt<<1;
pushdown(rt);
modify(target,nowl,nowm,rx,val); modify(target,nowm+1,nowr,rx|1,val);
}
int query(int target,int nowl,int nowr,int rt) {
if(nowl==nowr) {
// printf("FIND val:%d in rt:%d\n",tr[rt].lazy,rt);
return tr[rt].lazy;
}
int nowm=(nowl+nowr)>>1,rx=rt<<1;
pushdown(rt);
return nowm<target?query(target,nowm+1,nowr,rx|1):query(target,nowl,nowm,rx);
}
inline void work(signed CASE=1,bool FINAL_CASE=false) {
n=read(); q=read();
for(int i=1;i<=n;i++) {
a[i]=read();
}
build(1,n,1);
// for(int i=1;i<=n;i++) {
// printf("%d%c",query(i,1,n,1)," \n"[i==n]);
// }
while(q--) {
if(read()==1) {
int x=read(),y=read();
modify(x,1,n,1,y);
} else {
tr[1].lazy=max(tr[1].lazy,read());
}
}
for(int i=1;i<=n;i++) {
printf("%d%c",query(i,1,n,1)," \n"[i==n]);
}
return;
}
Day6 P464. 数数
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const int MAXN=(int)1e5+3;
int n,q;
int a[MAXN], b[MAXN], b_len;
struct ptree {
int val, l, r;
} tree[MAXN * 24];
int root[MAXN], node_cnt;
void modify(int pre, int l, int r, int& now, int pos) {
if(pos<l || pos>r) {
return;
}
tree[now = ++node_cnt] = {tree[pre].val + 1, tree[pre].l, tree[pre].r};
if (l == r) {
return;
}
int mid = (l+r)>>1;
modify(tree[pre].l, l, mid, tree[now].l, pos); modify(tree[pre].r, mid + 1, r, tree[now].r, pos);
}
int query(int l, int r, int nowl, int nowr, int k) {
int nowm=(nowl+nowr)>>1;
return (nowl==nowr) ? (k>=nowl ? tree[r].val-tree[l].val : 0) : ((nowm>k) ? query(tree[l].l,tree[r].l,nowl,nowm,k) : (nowm==k ? tree[tree[r].l].val-tree[tree[l].l].val : tree[tree[r].l].val-tree[tree[l].l].val+query(tree[l].r,tree[r].r,nowm+1,nowr,k)));
}
inline void work(signed CASE=1,bool FINAL_CASE=false) {
n=read(); q=read();
for(int i=1;i<=n;i++) {
b[i]=a[i]=read();
}
sort(b+1,b+n+1);
b_len = unique(b+1,b+n+1)-b-1;
for(int i=1;i<=n;i++) {
modify(root[i-1],1,b_len,root[i],lower_bound(b+1,b +b_len+1,a[i])-b);
}
for(int i=1;i<=q;i++) {
int l=read(),r=read(),h=read();
printf("%d%c",query(root[l-1],root[r],1,b_len,upper_bound(b+1,b+b_len+1,h)-b-1)," \n"[i==q]);
}
return;
}
signed main() {
// ios::sync_with_stdio(false); cin.tie(nullptr); cout.tie(nullptr); //freopen(".in", "r", stdin);//freopen(".out", "w", stdout);
signed T=(signed)read();//scanf("%d",&T);//cin>>T;
for(signed CASE=1; CASE<=T; CASE++) { //
//printf("Case #%d: ",CASE); //printf("Case %d: ",CASE); //printf("Case #%d: \n",CASE); //printf("Case %d: \n",CASE);
//while(cin>>n) work(n);
work(CASE,CASE==T);
if(CASE==T) {
break;
}
node_cnt=0; mmst0(root);
}
return 0;
}
Day7 P454. Minimum Or Spanning Tree
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const int MAXN=(int)4e5+3;
int n,m;
int fa[MAXN];
struct Edge {
int u,v,w;
} e[MAXN];
int findx(int x) {
return x==fa[x] ? x : fa[x]=findx(fa[x]);
}
bool merge(int u,int v) {
u=findx(u),v=findx(v);
return u==v ? false : fa[u]=v;
}
int chk(int mask) {
for(int i=1;i<=n;i++) fa[i] = i;
int size = n;
for(int i=1;i<=m;i++) {
if ((e[i].w | mask) == mask && merge(e[i].u, e[i].v)) {
size--;
}
}
return size == 1;
}
inline void work(signed CASE=1,bool FINAL_CASE=false) {
n=read(); m=read();
for(int u,v,w,i=1;i<=m;i++) {
e[i].u=read(); e[i].v=read(); e[i].w=read();
}
int ans=(1<<30)-1;
for(int i=29;i>=0;i--) {
ans ^= (1 << i);
if (!chk(ans)) ans ^= (1 << i);
}
printf("%lld\n",ans);
return;
}
//考虑每一位的贡献，从高到低用并查集判断在某一位全部为0时是否能够组成一棵树。如果可以，则禁用此位为1的所有边，继续向下查找。
//用并查集判断是否可以构成树（联通），如果所有节点的父亲节点相同，则图联通。
Day8 P466. 摘桃子
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const int MAXN=(int)2e5+3;
int n,k,ans;
int a[MAXN];
map<int,int> mp;
inline void work(signed CASE=1,bool FINAL_CASE=false) {
n=read(); k=read();
for(int i=1;i<=n;i++) {//先a[i] -= 1再前缀和跟先前缀和再把每个前缀和都 -i 等价，群友 NB
a[i]=(read()-1+a[i-1])%k;
}
// for(int i=1;i<=n;i++) printf("%lld%c",sum[i]," \n"[i==n]);
for(int i=0;i<=min(k-1,n);i++) { // for i in len
ans+=mp[a[i]];
mp[a[i]]++;
}
for(int i=min(k-1,n)+1;i<=n;i++) {
mp[a[i-k]]--;
ans+=mp[a[i]];
mp[a[i]]++;
}
// for(int i=1;i<=n;i++) {
// mp[sum[i]]++;
// for(int j=i;j<=n;j++) {
// // printf("DBG: %lld %lld SUM:%lld\n",i,j,sum[j]-sum[i-1]);
// if((sum[j]-sum[i-1]+k)%k==j-i+1) {
// ans++;
// // printf("ANS+: %lld %lld\n",i,j);
// }
// }
// }
printf("%lld\n",ans);
return;
}
Day9 P467. 路径计数2
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const int MAXN=(int)1e6+3,MOD=(int)1e9+7;
int n,m;
int x[MAXN],y[MAXN],dp[MAXN];
// 组合数板子
#define mod MOD
ll fac[MAXN<<1],inv[MAXN<<1];
ll pow_mod(ll a,ll n) {
ll ret =1;
while(n) {
if(n&1) ret=ret*a%mod;
a=a*a%mod;
n>>=1;
}
return ret;
}
void init() {
fac[0]=1;
for(int i=1;i<MAXN<<1;i++) {
fac[i]=fac[i-1]*i%mod;
}
}
ll Cc(ll x, ll y) {
return fac[x]*pow_mod(fac[y]*fac[x-y]%mod,mod-2)%mod;
}
#undef mod
// 组合数板子
int dfs(int i) {
if (dp[i]) return dp[i];
int res=0;
for (int j=1;j<=m+1;j++) {
if (j!=i && x[j]<=x[i] && y[j]<=y[i]) {
res=(res+(dfs(j)*Cc(x[i]-x[j]+y[i]-y[j],x[i]-x[j]))%MOD)%MOD;
}
}
return dp[i]=(Cc(x[i]-1+y[i]-1,x[i]-1)-res+MOD)%MOD;
}
inline void work(signed CASE=1,bool FINAL_CASE=false) {
n=read(); m=read();
for(int i=1;i<=m;i++) {
x[i]=read(); y[i]=read();
}
x[m+1]=y[m+1]=n;
dfs(m+1);
printf("%lld\n",dp[m+1]);
return;
}
signed main() {
init();
// ios::sync_with_stdio(false); cin.tie(nullptr); cout.tie(nullptr); //freopen(".in", "r", stdin);//freopen(".out", "w", stdout);
signed T=1;//(signed)read();//scanf("%d",&T);//cin>>T;
for(signed CASE=1; CASE<=T; CASE++) { //
//printf("Case #%d: ",CASE); //printf("Case %d: ",CASE); //printf("Case #%d: \n",CASE); //printf("Case %d: \n",CASE);
//while(cin>>n) work(n);
work(CASE,CASE==T);
}
return 0;
}
Day10 P468. 函数求和
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const int MAXN=(int)1e5+3,MOD=998244353;
int n,k,ans;
int a[MAXN];
int qpw(int a,int b) {
int res=1;
while(b) {
if(b&1) {
res=(res*a)%MOD;
}
b>>=1;
a=(a*a)%MOD;
}
return res;
}
inline void work(signed CASE=1,bool FINAL_CASE=false) {
n=read(); k=read(); bitset<64>b;
for(int i=1;i<=n;i++) {
bitset<64> a(read());
int cnt1=0,cnt0=0;
for(int j=0;j<k;j++) {
cnt1+=((!b[j]) && a[j]); // if(b[j]==0 && a[j]==1)cnt1++;
cnt0+=(!(b[j] || a[j])); // if(b[j]==0 && a[j]==0)cnt0++;
}
if(cnt1) {
ans=(ans+i*(qpw(2,cnt1)-1+MOD)%MOD*(qpw(2,cnt0)%MOD)%MOD)%MOD;
// 加入 a[i] 的贡献 ((2^cnt1-1)*(2^cnt2)个数会选中 a[i] )
}
b|=a; // mask掉已经用过的位
}
printf("%lld\n",ans);
return;
}
//没想出来，看了下面裙友题解才写出来的
// https://zhuanlan.zhihu.com/p/476870600
// a[i] , x , a[i]&x
// 0 , 0 , 0
// 0 , 1 , 0
// 1 , 0 , 0
// 1 , 1 , 1
// 仅在 a[i]==1 && x==0 时 a[i]&x！=a[i]
Day11 P497. XOR Inverse
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const int MAXN=(int)3e5+3;
int n,ans,x;
int a[MAXN],st[MAXN];
int cnt[40][2];
vector<int> vec;
void solve(vector<int> vec,int p) {
if(p==-1||vec.empty()) {
return;
}
vector<int> tmp[2];
for(auto x:vec) {
tmp[a[x]>>p&1].pb(x);
}
ll sum=0;
for(int i=0,j=0;i<sz(tmp[0]);i++) {
while(j<sz(tmp[1])&&tmp[0][i]>tmp[1][j]) ++j;
sum+=j;
}
cnt[p][0]+=sum;
cnt[p][1]+=1ll*sz(tmp[0])*sz(tmp[1])-sum;
solve(tmp[0],p-1); solve(tmp[1],p-1);
}
inline void work(signed CASE=1,bool FINAL_CASE=false) {
n=read();
for(int i=1;i<=n;i++) {
a[i]=read();
}
for(int i=1;i<=n;i++) {
vec.pb(i);
}
solve(vec,30);
for(int i=30;~i;i--) {
ans+=min(cnt[i][0],cnt[i][1]);
if(cnt[i][1]<cnt[i][0]) x|=1<<i;
}
printf("%lld %lld\n",ans,x);
return;
}
Day12 P469. Closest Equals
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const int MAXN=(int)5e5+3;
int n,m;
int a[MAXN],l[MAXN],r[MAXN],ans[MAXN],ord[MAXN],tr[MAXN];
map<int,int> mp;
inline void work(signed CASE=1,bool FINAL_CASE=false) {
// int j,k,t;
memset(tr,3,sizeof(tr));
n=read(); m=read();
for(int i=1;i<=n;i++) {
a[i]=read();
}
for(int i=0;i<m;i++) {
ord[i]=i;
l[i]=read(); r[i]=read();
}
sort(ord,ord+m,[](const int &xx,const int &yy) {
return r[xx]<r[yy];
});
for(int i=0,j=1;i<m;i++) {
int t=ord[i];
while(j<=r[t]) {
if(mp.count(a[j])) {
for(int k=mp[a[j]];k;k-=k&-k) {
tr[k]=min(tr[k],j-mp[a[j]]);
}
}
mp[a[j]]=j;
j++;
}
ans[t]=1e6;
for(int k=l[t];k<=n;k+=k&-k) {
ans[t]=min(ans[t],tr[k]);
}
}
for(int i=0;i<m;i++) {
printf("%d\n",ans[i]==1e6?-1:ans[i]);
}
return;
}
Day13 P380. CCPC Harbin 2021 G, Damaged Bicycle
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const int MAXN=(int)1e6+3;
struct node{int to,w;};
vector <node> mp[MAXN];
inline bool operator < (node a,node b){return a.w>b.w;}
inline bool operator > (node a,node b){return a.w<b.w;}
priority_queue <node> q;
int n,m,t,r,k,A[MAXN],d[25][MAXN],id[MAXN],dp[MAXN][25],ct[MAXN];
bool vis[25][MAXN];
double DP[MAXN][25],P[MAXN];
inline void dij(int id,int uu) {
for(int i=1;i<=n;i++) {
d[id][i]=INF;
}
q.push({uu,0});
d[id][uu]=0;
while(!q.empty()) {
node u=q.top(); q.pop();
if(vis[id][u.to]) {
continue;
}
d[id][u.to]=u.w; vis[id][u.to]=1;
for(auto v : mp[u.to]) {
if(d[id][v.to]>d[id][u.to]+v.w) {
d[id][v.to]=d[id][u.to]+v.w;
q.push({v.to,d[id][v.to]});
}
}
}
}
inline double DFS(int sta,int u) {
if(DP[sta][u]) return DP[sta][u];
double tmp=1.0*P[u]*d[u][n]/t+(1-P[u])*d[u][n]/r;
for(int i=1;i<=k;i++) {
if(sta&(1<<(i-1))) continue;
tmp=min(tmp,1.0*(1-P[u])*d[u][n]/r+P[u]*(1.0*d[u][A[i]]/t+DFS(sta|(1<<(i-1)),i)));
}
return DP[sta][u]=tmp;
}
inline void work(signed CASE=1,bool FINAL_CASE=false) {
t=read(); r=read(); n=read(); m=read();
for(int i=1;i<=m;i++) {
int u=read(),v=read(),w=read();
mp[u].pb((node){v,w}); mp[v].pb((node){u,w});
}
k=read();
for(int i=1;i<=k;i++) A[i]=read(),P[i]=read()/100.0,dij(i,A[i]),id[A[i]]=i;
dij(19,1); dij(20,n); P[19]=1;
if(d[19][n]==INF) {
puts("-1");
return;
}
DFS(0,19);
int ed=(1<<k)-1;
printf("%.6f\n",DP[0][19]);
return;
}
Day14 P501. 拆方块
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const int MAXN=(int)1e5+3;
int n,ans;
int l[100005],r[100005],h[100005];
inline void work(signed CASE=1,bool FINAL_CASE=false) {
n=read()+1;
for(int i=1;i<n;i++) {
h[i]=read();
}
for(int i=1;i<n;i++) {
l[i]=l[i-1]+1;
if(l[i]>h[i]) {
l[i]=h[i];
}
}
for(int i=n-1;i;i--) {
r[i]=r[i+1]+1;
if(r[i]>h[i]) {
r[i]=h[i];
}
}
for(int i=1;i<n;i++) {
ans=max(ans,min(l[i],r[i]));
}
printf("%d\n",ans);
return;
}
Day15 <a class="link" href="http://oj.daimayuan.top/problem/504" target="_blank" rel="noopener"
>P504. 连续子序列
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const int MAXN=(int)2e5+3;
int n,ans,ansi;
map<int,int> f;
inline void work(signed CASE=1,bool FINAL_CASE=false) {
n=read();
for(int x,i=1;i<=n;i++) {
f[x]=max(f[x],f[(x=read())-1]+1);
}
for(auto x:f) {
if(x.se>ans) {
ans=x.se;
ansi=x.fi;
}
}
printf("%lld\n",ans);
for(int i=ansi-ans+1;i<=ansi;i++) {
printf("%d%c",i," \n"[i==ansi]);
}
return;
}
Day16 <a class="link" href="http://oj.daimayuan.top/problem/503" target="_blank" rel="noopener"
>P503. 工作安排
正解写不对遂弃疗
原题找了个树状数组的写法
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#include<bits/stdc++.h>
#define ll long long
using namespace std;
inline int read()
{
char ch=getchar();
int res=0,flag=1;
while(ch<'0'||ch>'9')
{
if(ch=='-')
flag=-1;
ch=getchar();
}
while(ch>='0'&&ch<='9')
{
res=res*10+ch-'0';
ch=getchar();
}
return res*flag;
}
int n;
struct work
{
int w;
int t;
bool operator <(work x2)
{
return w>x2.w;
}
}x[100001];
int c[100001];
ll ans;
inline int lowbit(int a)
{
return a&(-a);
}
inline void update(int pos,int v)
{
while(pos<=n)
{
c[pos]+=v;
pos+=lowbit(pos);
}
}
inline int sum(int pos)
{
int res=0;
while(pos)
{
res+=c[pos];
pos-=lowbit(pos);
}
return res;
}
int main()
{
n=read();
for(int i=1;i<=n;++i)
{
x[i].t=min(read(),(int)1e5);
x[i].w=read();
}
sort(x+1,x+n+1);
for(int i=1;i<=n;++i)
if(sum(x[i].t)!=x[i].t)
{
int l=1,r=x[i].t,mid,res;
while(l<=r)
{
mid=(l+r)>>1;
if(sum(x[i].t)-sum(mid-1)!=x[i].t-mid+1)
{
res=mid;
l=mid+1;
}
else
r=mid-1;
}
update(res,1);
ans+=x[i].w;
}
printf("%lld\n",ans);
return 0;
}
Day17 <a class="link" href="http://oj.daimayuan.top/problem/505" target="_blank" rel="noopener"
>P505. 三角果计数
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const int MAXN=(int)1e5+3;
int n,ans;
int len[MAXN];
vector<int> grap[MAXN];
void dfs(int u, int fa) {
for (auto v : grap[u]) {
if (v != fa) {
dfs(v, u);
ans += len[u] * len[v] * (n - len[v] - len[u] - 1);
len[u] += len[v];
}
}
len[u]++;
}
inline void work(signed CASE=1,bool FINAL_CASE=false) {
n=read();
for (int u,v,w,i = 1; i < n; i ++) {
grap[u=read()].push_back(v=read());
grap[v].push_back(u);
w=read();
}
dfs(1, INF);
printf("%lld\n",ans);
return;
}
//https://zhuanlan.zhihu.com/p/480494165
Day18 <a class="link" href="http://oj.daimayuan.top/problem/555" target="_blank" rel="noopener"
>P555. 整齐的数组2
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const int MAXN=(int)2e5+3;
int n;
int a[43];
inline void work(signed CASE=1,bool FINAL_CASE=false) {
n=read();
for(int i=1;i<=n;i++) {
a[i]=read();
}
sort(a+1,a+1+n);
int cnt = 1;
for(int i=2;i<=n;i++) {
if(a[i]==a[i-1]) cnt++;
else cnt = 1;
if(cnt >= n/2) {
cout<<"-1\n";
return ;
}
}
int res = 0;
for(int i=1;i<=n;i++) {
map<int ,int >mp;
cnt = 0;
for(int j=1;j<=n;j++) {
int x = abs(a[j]-a[i]);
if(a[i]==a[j]) {
cnt ++ ;
continue;
}
for(int k=1;k*k<=x;k++) {
if(x % k==0) {
if(k*k==x) mp[k]++;
else mp[x/k]++,mp[k]++;
}
}
}
for(auto i : mp) {
if(i.se + cnt >= n/2) {
res = max(res,i.fi);
}
}
}
printf("%d\n",res);
return;
}
signed main() {
// ios::sync_with_stdio(false); cin.tie(nullptr); cout.tie(nullptr); //freopen(".in", "r", stdin);//freopen(".out", "w", stdout);
signed T=(signed)read();//scanf("%d",&T);//cin>>T;
for(signed CASE=1; CASE<=T; CASE++) { //
//printf("Case #%d: ",CASE); //printf("Case %d: ",CASE); //printf("Case #%d: \n",CASE); //printf("Case %d: \n",CASE);
//while(cin>>n) work(n);
work(CASE,CASE==T);
}
return 0;
}
Day19 <a class="link" href="http://oj.daimayuan.top/problem/556" target="_blank" rel="noopener"
>P556. 三进制循环
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const int MAXN=(int)5e5+3;
vector<int> grap[MAXN];
int n,ans;
int a[MAXN];
int dp[MAXN][2];
void dfs(int u,int fa) {
dp[u][0]=dp[u][1]=1; // 初始赋值1
for (int v : grap[u]) {
if(v==fa) {
continue;
}
dfs(v, u);
if((a[u]+1)%3==a[v]) {
dp[u][0]=max(dp[u][0],dp[v][0]+1); // 递增
}
if((a[v]+1)%3==a[u]) {
dp[u][1]=max(dp[u][1],dp[v][1]+1); // 递减
}
}
ans = max(ans, dp[u][0] + dp[u][1] - 1);//记得减1
}
inline void work(signed CASE=1,bool FINAL_CASE=false) {
n=read();
for(int u,v,i=1;i<n;i++) {
grap[u=read()].pb(v=read());
grap[v].pb(u);
}
for(int i=1;i<=n;i++) {
a[i]=read();
}
dfs(1,0);
printf("%lld\n",ans);
return;
}
// https://zhuanlan.zhihu.com/p/481678880 严格鸽，我滴超人
Day20 <a class="link" href="http://oj.daimayuan.top/problem/559" target="_blank" rel="noopener"
>P559. 树上逆序对
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const int MAXN=(int)2e5+3;
int n;
int s[MAXN],res[MAXN];
PII a[MAXN];
void update(int x) {
for(int i=x;i<=n;i+=lowbit(i)) {
s[i]++;
}
}
int query(int x,int res=0){
for (int i=x;i!=0;i-=lowbit(i)) {
res+=s[i];
}
return res;
}
inline void work(signed CASE=1,bool FINAL_CASE=false) {
n=read();
for(int i=1;i<=n;i++) {
a[i].fi=read(),a[i].se = i;
}
sort(a+1,a+1+n);
for(int i=1;i<=n;i++) {
for (int j=1;j<n;j++) {
int l=(a[i].se-1)*j+2,r=min(j*a[i].se+1,n);
if(l>r) {
break;
}
res[j]+=(query(r)-query(l-1));
}
update(a[i].se);
}
for(int i=1;i<n;i++) {
printf("%d%c",res[i]," \n"[i==(n-1)]);
}
return;
}
Day21 <a class="link" href="http://oj.daimayuan.top/problem/560" target="_blank" rel="noopener"
>P560. 约分
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const int MAXN=(int)1e5+3;
int ans1,ans2,r1,r2;
string s,t;
int cnt[10];
bool find(string &s,string &t)
{
int n=sz(s),m=sz(t),i=0;
for(auto c:t)
{
while (i<n&&s[i]!=c) {
cnt[s[i++]-'0']--;
}
if (i==n) {
return false;
}
i++;
}
while(i<n) {
cnt[s[i++]-'0']--;
}
for (int i=0;i<10;i++) {
if (cnt[i]) {
return false;
}
}
return true;
}
void work(int CASE,bool FINAL_CASE)
{
cin>>s>>t;
int n=sz(s),m=sz(t);
for (auto &x:s) {
r1=r1*10+(x-'0');
}
for (auto &x:t) {
r2=r2*10+(x-'0');
}
int ans1=r1,ans2=r2,now=__gcd(r1,r2);
r1/=now; r2/=now;
for (int j=0;j<1<<n;j++)
{
int r3=0,ctt=0;
mmst0(cnt);
for (int i=0;i<n;i++) {
if (j>>i&1) {
r3=r3*10+s[i]-'0';
} else {
cnt[s[i]-'0']++;
ctt++;
}
}
if (r3%r1||!r3||(__int128)r3/r1*r2>=1llu<<63) {
continue;
}
ll r4=r3/r1*r2;
string tar=to_string(r4);
while (tar.size()+ctt<m) {
tar='0'+tar;
}
if (r3<ans1&&find(t,tar)) {
ans1=r3,ans2=r4;
}
}
cout<<ans1<<' '<<ans2<<endl;
}
signed main() {
ios::sync_with_stdio(false); cin.tie(nullptr); cout.tie(nullptr); //freopen(".in", "r", stdin);//freopen(".out", "w", stdout);
signed T=1;//(signed)read();//scanf("%d",&T);//cin>>T;
for(signed CASE=1; CASE<=T; CASE++) { //
//printf("Case #%d: ",CASE); //printf("Case %d: ",CASE); //printf("Case #%d: \n",CASE); //printf("Case %d: \n",CASE);
//while(cin>>n) work(n);
work(CASE,CASE==T);
}
return 0;
}
//nmd,wsm 又是一个我在赛场上没出的题，当时这题甚至没开没看题意
Day22 <a class="link" href="http://oj.daimayuan.top/problem/562" target="_blank" rel="noopener"
>P562. 蜗蜗的数列
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const int MAXN=(int)1e6+3;
int n,q,mod,cnt;
int v[MAXN],fib[MAXN];
void upd(int pos,int val){
if(pos<n && v[pos]==0) {
cnt--;
}
v[pos]=(v[pos]+val)%mod;
if(pos<n && v[pos]==0) {
cnt++;
}
}
inline void work(signed CASE=1,bool FINAL_CASE=false) {
cnt=n=read(); q=read(); mod=read();
fib[0]=fib[1]=1%mod;
for(int i=2;i<MAXN;i++) {
fib[i]=(fib[i-1]+fib[i-2])%mod;
}
for(int a,i=0;i<n;i++) {
a=read();
upd(i,a); upd(i+1,-a); upd(i+2,-a);
}
for(int b,i=0;i<n;i++) {
b=read();
upd(i,-b); upd(i+1,b); upd(i+2,b);
}
for(int l,r,val,i=1;i<=q;i++) {
char c=nc(); l=read()-1; r=read()-1;
val=(c=='A'?1:-1);
upd(l,val*fib[0]); upd(r+1,-val*fib[r+1-l]); upd(r+2,-val*fib[r-l]);
puts(n==cnt?"Yes":"No");
}
return;
}
Day23 <a class="link" href="http://oj.daimayuan.top/problem/131" target="_blank" rel="noopener"
>P131. 最大公约数
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const int MAXN=(int)2e3+3;
int n,m,mx1,sum;
int a[MAXN],mx[MAXN],t[MAXN*10],b[MAXN];;
void calc(int q,int mo) {
sum=0,mx1=1;
for(int j=1;j<=n;j++) {
b[j]=a[j]-a[j]/mo*mo;
}
sort(b+1,b+1+n);
for(int j=1;j<=q;j++) {
if(b[j]!=b[j-1]) {
mx1=max(mx1,sum);
sum=1;
} else {
sum++;
}
}
mx[mx1]=max(mx[mx1],mo);
}
inline void work(signed CASE=1,bool FINAL_CASE=false) {
n=read();
for(int i=1;i<=n;i++) {
a[i]=a[i-1]+read();
}
mx[1]=m=a[n];
b[0]=b[n+1]=-1;
for(int i=1;i<=sqrt(m);i++) {
if(m%i==0) {
calc(n+1,i);
calc(n,m/i);
}
}
for(int i=n;i>=1;i--) {
mx[i]=max(mx[i],mx[i+1]);
}
for(int i=1;i<=n;i++) {
printf("%lld\n",mx[i]);
}
return;
}
Day24 <a class="link" href="http://oj.daimayuan.top/problem/607" target="_blank" rel="noopener"
>P607. 平方计数
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const int MAXN=(int)1e6+3;
int n,ans;
int a[MAXN],cot[MAXN];
inline void work(signed CASE=1,bool FINAL_CASE=false) {
n=read();
for (int i=1;i<=n;i++) {
cot[a[i]=read()]++;
}
for (int i=1;i<=MAXN;i++) {
for (int j=i;j<=MAXN;j+=i) {
int mi=min(j/i,i),mx=max(j/i,i);
int d=mx-mi;
if (!(d%2)) {
ans+=cot[j]*cot[d>>1];
}
}
}
printf("%lld\n",ans>>1);
return;
}
Day25 <a class="link" href="http://oj.daimayuan.top/problem/608" target="_blank" rel="noopener"
>P608. 字典序最小
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const int MAXN=(int)1e5+3;
int n,k,cnt;
bool instk[MAXN];
int last[MAXN],a[MAXN],ans[MAXN];
stack<int> stk;
inline void work(signed CASE=1,bool FINAL_CASE=false) {
n=read(); k=read();
for (int i = 1; i <= n; i++) {
last[a[i]=read()]=i;
}
for (int i = 1; i <= n; i++) {
if(!instk[a[i]]) {
while (stk.size() && stk.top() > a[i] && last[stk.top()] > i) {
instk[stk.top()]=false;
stk.pop();
}
stk.push(a[i]);
instk[a[i]]=true;
}
}
while (!stk.empty()) {
ans[++cnt]=stk.top();
stk.pop();
}
for(int i=cnt;i>=1;i--) {
printf("%d%c",ans[i]," \n"[i==1]);
}
return;
}
Day26 <a class="link" href="http://oj.daimayuan.top/problem/611" target="_blank" rel="noopener"
>P611. 拆拆
不会，只拿了42分，弃疗
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//2020420 贴一个不是我的代码，这位帮我 debug 后过了
const int N=2e6+5;
int pre[N],vis[N];
int cnt=1;
void get_pri() {
f(i,2,1e6+5) {
if(!vis[i])pre[cnt++]=i;
for(int j=i+i;j<=1e6+6;j=j+i) vis[j]=1;
}
}
int x,k;
int ans[N];
const int mod=1e9+7;
int power(int a,int b) {
int ans1=1;
while(b) {
if(b&1) ans1=ans1*a%mod;
a=a*a%mod;
b>>=1;
}
return ans1;
}
int jc[N];
void get_jc() {
int now=1;
jc[0]=1;
f(i,1,1e6)
{
now=now*i%mod;
jc[i]=now;
}
}
int C(int a,int b) {
return jc[a]*power(jc[a-b],mod-2)%mod*power(jc[b],mod-2)%mod;
}
int p[N],m,c[N];
void divide(int n) {
m=0;
f(i,2,sqrt(n)) {
if(n%i==0) {
p[++m]=i,c[m]=0;
while(n%i==0)n/=i,c[m]++;
}
}
if(n>1)p[++m]=n,c[m]=1;
}
inline void solve() {
scanf("%lld%lld",&x,&k);
int temp=x;
divide(x);
int ansi=1;
f(i,1,m) {
if(!c[i])continue;
ansi=(ansi*C(c[i]+k-1,k-1))%mod;
}
ansi=ansi*power(2,k-1)%mod;
printf("%lld\n",ansi);
return;
}
Day27 <a class="link" href="http://oj.daimayuan.top/problem/614" target="_blank" rel="noopener"
>P614. “Z”型矩阵
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const int MAXN=(int)3e3+3;
int n,m,ans;
bitset<MAXN> a[MAXN], b[MAXN], c[MAXN];
char s[MAXN];
inline void work(signed CASE=1,bool FINAL_CASE=false) {
// n=read(); m=read();
scanf("%lld%lld",&n,&m);
for(int i = 0; i < n; i++){
scanf("%s", s);
for(int j = 0; j < m; j++)
a[i][j] = (s[m-1-j] == 'z');
b[i].set(); c[i].set();
}
for(int k = 0; k < m && k < n; k++){
for(int i = 0; i < n; i++)
b[i] &= a[i] >> k;
for(int i = 0; i + k < n; i++){
c[i] &= a[i+k] >> k;
ans += (b[i] & b[i+k] & c[i]).count();
}
}
printf("%lld\n",ans);
return;
}
Day28 <a class="link" href="http://oj.daimayuan.top/problem/613" target="_blank" rel="noopener"
>P613. 好序列
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const int MAXN=(int)2e5+3;
int n;
int a[MAXN],pre[MAXN],nxt[MAXN];
bool chk(int l,int r) {
if(l>=r) {
return true;
}
int x=l,y=r;
while(x<=y) { // 启发式合并
if(pre[x]<l && r<nxt[x]) {
return (chk(l,x-1) && chk(x+1,r));
}
x++;
if(pre[y]<l && r<nxt[y]) {
return (chk(l,y-1) && chk(y+1,r));
}
y--;
}
return false;
}
inline void work(signed CASE=1,bool FINAL_CASE=false) {
n=read(); map<int,int> mp;
for(int i=1;i<=n;i++) {
pre[i]=-1; nxt[i]=n+1;
}
for(int i=1;i<=n;i++) {
a[i]=read();
}
for(int i=1;i<=n;i++) {
pre[i]=mp[a[i]];
nxt[mp[a[i]]]=i;
mp[a[i]]=i;
}
if(chk(1,n)) {
YESS;
} else {
NOO;
}
return;
}
signed main() {
// ios::sync_with_stdio(false); cin.tie(nullptr); cout.tie(nullptr); //freopen(".in", "r", stdin);//freopen(".out", "w", stdout);
signed T=(signed)read();//scanf("%d",&T);//cin>>T;
for(signed CASE=1; CASE<=T; CASE++) { //
//printf("Case #%d: ",CASE); //printf("Case %d: ",CASE); //printf("Case #%d: \n",CASE); //printf("Case %d: \n",CASE);
//while(cin>>n) work(n);
work(CASE,CASE==T);
if(CASE!=T) { }
}
return 0;
}
Day29 <a class="link" href="http://oj.daimayuan.top/problem/606" target="_blank" rel="noopener"
>P606. 社交圈
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const int MAXN=(int)1e5+3;
int n,ans;
int l[MAXN],r[MAXN];
inline void work(signed CASE=1,bool FINAL_CASE=false) {
n=read();
for(int i=1;i<=n;i++) {
l[i]=read(); r[i]=read();
}
sort(l+1,l+1+n);
sort(r+1,r+1+n);
for(int i=1;i<=n;i++) {
ans+=max(l[i],r[i]);
}
printf("%lld\n",ans+n); // +n 包括自己的椅子
return;
}
// 草，是若干个圈
// 开始以为只有一个圈
Day30 <a class="link" href="http://oj.daimayuan.top/problem/609" target="_blank" rel="noopener"
>P609. 区间和
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const int MAXN=(int)2e5+3;
int n,q;
int fa[MAXN];
int findx(int x) {
return x==fa[x]?x:fa[x]=findx(fa[x]);
}
void merge(int x,int y) {
fa[findx(x)]=findx(y);
}
inline void work(signed CASE=1,bool FINAL_CASE=false) {
n=read(); q=read();
for(int i=1;i<=n;i++) {
fa[i]=i;
}
while(q--) {
int l=read(),r=read();
merge(l-1,r);
}
if(findx(0)==findx(n)) {
YESS;
} else {
NOO;
}
return;
}
Day31 <a class="link" href="http://oj.daimayuan.top/problem/618" target="_blank" rel="noopener"
>P618. 选数2
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const int MAXN=(int)2e5+3;
int n,p,q,maxx,cans=0;
int pre[MAXN],ans[MAXN];
PII a[MAXN];
// ===================================================================================================================
// Seg Tree 板子
struct Node{
int val,add,mul=1;
} t[MAXN<<2];
// int a[100003];
void build_t(int root,int l,int r)
{
t[root].mul=1,t[root].add=0;
if(l==r) t[root].val=0; //本题修改的地方,val=0 打标记用
else
{
int m=(l+r)>>1,rx=root<<1;
build_t(rx,l,m);
build_t(rx+1,m+1,r);
t[root].val=(t[rx].val+t[rx+1].val);
}
}
void push_down(int root,int l,int r)
{
int m=(l+r)>>1,rx=root<<1;
t[rx].val=(t[rx].val*t[root].mul+t[root].add*(m-l+1));
t[rx+1].val=(t[rx+1].val*t[root].mul+t[root].add*(r-m));//!!!!!!!!!!!!!!!!!!rx+1
t[rx].mul=(t[rx].mul*t[root].mul);
t[rx+1].mul=(t[rx+1].mul*t[root].mul);
t[rx].add=(t[rx].add*t[root].mul+t[root].add);
t[rx+1].add=(t[rx+1].add*t[root].mul+t[root].add);
t[root].mul=1,t[root].add=0;
}
void up1(int root,int nowl,int nowr,int l,int r,ll k)
{
if(r<nowl || nowr<l) return;
if(l<=nowl && nowr<=r)
{
t[root].val=(t[root].val*k);
t[root].mul=(t[root].mul*k);
t[root].add=(t[root].add*k);
return;
}
push_down(root,nowl,nowr);
int m=(nowl+nowr)>>1,rx=root<<1;
up1(rx,nowl,m,l,r,k);
up1(rx+1,m+1,nowr,l,r,k);
t[root].val=(t[rx].val+t[rx+1].val);
}
void up2(int root,int nowl,int nowr,int l,int r,ll k)
{
if(r<nowl || nowr<l) return;
if(l<=nowl && nowr<=r)
{
t[root].add=(t[root].add+k);
t[root].val=(t[root].val+k*(nowr-nowl+1));
return;
}
push_down(root,nowl,nowr);
int m=(nowl+nowr)>>1,rx=root<<1;
up2(rx,nowl,m,l,r,k);
up2(rx+1,m+1,nowr,l,r,k);
t[root].val=(t[rx].val+t[rx+1].val);
}
ll query(int root,int nowl,int nowr,int l,int r)
{
if(r<nowl || nowr<l) return 0;
if(l<=nowl && nowr<=r) return t[root].val;
push_down(root,nowl,nowr);
int m=(nowl+nowr)>>1,rx=root<<1;
return (query(rx,nowl,m,l,r)+query(rx+1,m+1,nowr,l,r));
}
// End Seg Tree
// ============================================================================================
int que(int l,int r) {
return l>r?0:pre[r]-pre[l-1];
}
bool chk(int m) {
for(int i=m;i<=n;i++) {
int mx=a[i].fi;
if(mx*q*m<=p*que(i-m+1,i)) {
return true;
}
}
return false;
}
int cal(int mx, int sum) {
return mx * q*maxx <= p * sum;
}
inline void work(signed CASE=1,bool FINAL_CASE=false) {
n=read();
for(int i=1;i<=n;i++) {
a[i].fi=read(); a[i].se=i;
}
p=read(); q=read();
sort(a+1,a+1+n);
for(int i=1;i<=n;i++) {
pre[i]=pre[i-1]+a[i].fi;
}
int l=0,r=n;
while(l<r) {
int m=(l+r)>>1;
if(chk(m)) {
maxx=max(maxx,m);
l=m+1;
} else {
r=m;
}
}
build_t(1,1,n);
for (int i = maxx; i <= n; i++) {
if (cal(a[i].fi, que(i - maxx + 1, i))) {
//[i - m + 1 , i ] 这个区间是满足条件的，我们才开始二分
int L = 1, R = i - maxx + 1, pos = INF;
int sum = que(i - maxx + 1 + 1, i);//这里记录下 i 左边 m - 1 个数的和
// for (int j = 1; j <= 50; j++) {
while(L<=R) {
int mid = (L + R) / 2;
if (cal(a[i].fi , sum + a[mid].fi)) {//最大的数是 a[i].val
pos = min(mid, pos);
R = mid-1;
}
else {
L = mid+1;
}
}
up2(1,1,n,pos,i,1);
}
}
for (int i = 1; i <= n; i++) {
if (query(1,1,n,i,i) == 0) {
ans[++cans]=a[i].se;
}
}
sort(ans+1, ans+1+cans);
printf("%lld\n",cans);
for (int i=1;i<=cans;i++) {
printf("%lld%c",ans[i]," \n"[i==cans]);
}
return;
}
Day32 <a class="link" href="http://oj.daimayuan.top/problem/665" target="_blank" rel="noopener"
>P665. 数组划分
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const int MAXN=(int)1e2+3;
int n,k,ans;
int a[MAXN],sum[MAXN];
bool chk(int x) { // 问是否存在 $k$ 段连续的子数组满足其按位与的结果 $\geqslant x$
bitset<MAXN> dp[MAXN];
dp[0][0]=1;
for(int i=1;i<=n;i++) { // $n \leqslant 100$
for(int j=1;j<=i;j++) {
if(((sum[i]-sum[j-1])&x)==x) { // x(BIN) 为 1 的位一定 1 x(BIN) 为 0 的地方 01 任意,所以可能存在大于等于当前 x 的
for(int l=1;l<=k;l++) { // l->len
dp[i][l]=(dp[i][l])|(dp[j-1][l-1]); // dp[前 i 个数][划分成 l 个子数组] = 其 & 的结果是否可能存在大于 x
}
}
}
}
return dp[n][k];
}
inline void work(signed CASE=1,bool FINAL_CASE=false) {
n=read(); k=read();
for(int i=1;i<=n;i++) {
sum[i]=sum[i-1]+(a[i]=read());
}
for(int i=60;i>=0;i--) {
ans|=(1LL<<i);
if(!chk(ans)) {
ans^=(1LL<<i);
}
}
printf("%lld\n",ans);
return;
}
// 划分数组再 & 后最大值也就是说 位置相同的 1 最多
// nmd , bitset 不能单纯对某一位 |=
Day33 <a class="link" href="http://oj.daimayuan.top/problem/678" target="_blank" rel="noopener"
>P678. namonamo
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const int MAXN=(int)1e5+3;
int n;
string s;
set<string>sst;
bool flag;
void dfs1(int st, int ed, string &a, string &b) {
if (st > ed) {
for (int i = 0; i < min(sz(a), sz(b)); i++) {
if (a[i] != b[i]) {
return;
}
}
if (sz(a)> sz(b)) {
sst.insert(a.substr(sz(b)));
} else {
sst.insert(b.substr(sz(a)));
}
return;
}
a.push_back(s[st]); dfs1(st + 1, ed, a, b); a.pop_back();
b.push_back(s[st]); dfs1(st + 1, ed, a, b); b.pop_back();
}
void cal(string a, string b) {
reverse(a.begin(), a.end()); reverse(b.begin(), b.end());
for (int i = 0; i < min(sz(a), sz(b)); i++) {
if (a[i] != b[i]) {
return;
}
}
string pre;
if (sz(a)> sz(b)) {
pre = (a.substr(sz(b)));
} else {
pre = (b.substr(sz(a)));
}
reverse(pre.begin(), pre.end());
if (sst.count(pre)) {
flag = true;
}
}
void dfs2(int st, int ed, string &a, string &b) {
if (flag) {
return;
}
if (st > ed) {
cal(a, b);
return;
}
a.push_back(s[st]); dfs2(st + 1, ed, a, b);
a.pop_back(); b.push_back(s[st]);
dfs2(st + 1, ed, a, b); b.pop_back();
}
inline void work(signed CASE=1,bool FINAL_CASE=false) {
sst.clear(); flag=false;
string tmp; cin >> tmp;
n = tmp.size(); s = "?" + tmp;
string a = "", b = "";
dfs1(1, n / 2, a, b);
dfs2(n / 2 + 1, n, a, b);
if (flag) {
YESS;
} else {
NOO;
}
return;
}
signed main() {
ios::sync_with_stdio(false); cin.tie(nullptr); cout.tie(nullptr); //freopen(".in", "r", stdin);//freopen(".out", "w", stdout);
signed T=1;//(signed)read();//scanf("%d",&T);//
cin>>T;
for(signed CASE=1; CASE<=T; CASE++) { //
//printf("Case #%d: ",CASE); //printf("Case %d: ",CASE); //printf("Case #%d: \n",CASE); //printf("Case %d: \n",CASE);
//while(cin>>n) work(n);
work(CASE,CASE==T);
if(CASE!=T) {}
}
return 0;
}
Day34 <a class="link" href="http://oj.daimayuan.top/problem/668" target="_blank" rel="noopener"
>P668. 体育节
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const int MAXN=(int)2e3+3;
int n;
int a[MAXN],dp[MAXN][MAXN];
inline void work(signed CASE=1,bool FINAL_CASE=false) {
n=read();
for(int i=1;i<=n;i++) {
a[i]=read();
}
sort(a+1,a+1+n);
for(int l=2;l<=n;l++) {
for(int i=1;i<=n-l+1;i++) {
int j=i+l-1;
dp[i][j]=a[j]-a[i]+min(dp[i+1][j],dp[i][j-1]);
}
}
printf("%lld\n",dp[1][n]);
return;
}
Day35 <a class="link" href="http://oj.daimayuan.top/problem/670" target="_blank" rel="noopener"
>P670. 测温
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const int MAXN=(int)1e6+3;
int n,ans,cnt=1,cur=-INF;
PII a[MAXN];
deque<int> q;
inline void work(signed CASE=1,bool FINAL_CASE=false) {
n=read();
for(int i=1;i<=n;i++) {
a[i].fi=read(); a[i].se=read();
while(!q.empty() && a[q.front()].fi > a[i].se) {
q.pop_front();
}
if(!q.empty()) {
ans=max(ans,i-q.front()+1);
}
int t=i;
while((!q.empty()) && a[i].fi > a[q.back()].fi) {
t=q.back(); q.pop_back();
}
a[t].fi=a[i].fi; q.push_back(t);
}
printf("%lld\n",ans);
return;
}
Day36 <a class="link" href="http://oj.daimayuan.top/problem/699" target="_blank" rel="noopener"
>P699. 并行排序
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const int MAXN=(int)1e6+3;
int n,len;
int a[MAXN],dp[MAXN];
inline void work(signed CASE=1,bool FINAL_CASE=false) {
n=read(),len=1;
for(int i=0;i<n;i++) {
a[i]=read();
}
dp[1]=a[0];
for(int i=1;i<n;i++) {
int l=0,r=len;
while(l<r) {
int m=(l+r+1)>>1;
if(dp[m]>a[i]) {
l=m;
} else {
r=m-1;
}
}
len=max(len,r+1);
dp[r+1]=a[i];
}
printf("%lld\n",len);
return;
}
signed main() {
// ios::sync_with_stdio(false); cin.tie(nullptr); cout.tie(nullptr); //freopen(".in", "r", stdin);//freopen(".out", "w", stdout);
signed T=(signed)read();//scanf("%d",&T);//cin>>T;
for(signed CASE=1; CASE<=T; CASE++) { //
//printf("Case #%d: ",CASE); //printf("Case %d: ",CASE); //printf("Case #%d: \n",CASE); //printf("Case %d: \n",CASE);
//while(cin>>n) work(n);
work(CASE,CASE==T);
if(CASE!=T) {}
}
return 0;
}
Day37 <a class="link" href="http://oj.daimayuan.top/problem/702" target="_blank" rel="noopener"
>P702. 丹钓战
令 $dp_{i,j}$ 表示从 $i$ 开始入栈，第 $2^j$ 个 成功的二元组的编号。
通过一遍对题意得模拟我们可以求出对于所有的 $i$ 的 $dp_{i,0}$ 。即当 $i$ 把 $j$ 弹出时，记录 $dp_{j,0}=i$
接着，预处理倍增数组，在 $O(n\log n)$ 的时间内求出所有的 $dp_{i,j}$
对于每次询问，枚举 $i$ 从 $\log n$ 到 $0$，看当前位往后 $2^i$ 个成功的二元组的编号是否还在 $[l,r]$ 区间内（类似倍增求 LCA）。单次询问时间复杂度 $\log n$
时间复杂度 $O(n\log n+q\log n)$
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const int MAXN=(int)5e5+3;
int n,q;
int a[MAXN],b[MAXN];
int dp[MAXN][23];
PII p[MAXN];
// stack<PII> stk;
stack<int> stk;
// int BF(int l,int r) { // Just for test
// int cnt=0;
// stack<PII> s;
// for(int i=l;i<=r;i++) {
// while(s.size() && (p[i].fi==s.top().fi || p[i].se>=s.top().se)) {
// s.pop();
// }
// s.push(p[i]);
// if(s.size()==1) {
// cnt++;
// }
// }
// return cnt;
// }
inline void work(signed CASE=1,bool FINAL_CASE=false) {
n=read(); q=read();
for(int i=1;i<=n;i++) {
p[i].fi=read();
}
for(int i=1;i<=n;i++) {
p[i].se=read();
}
for(int i=1;i<=n+1;i++) {
dp[i][0]=n+1;
}
for(int i=1;i<=n;i++) {
while(stk.size() && (p[i].fi==p[stk.top()].fi || p[i].se>=p[stk.top()].se)) {
dp[stk.top()][0]=i;
stk.pop();
}
stk.push(i);
}
for(int j=1;j<=20;j++) {
for(int i=1;i<=n+1;i++) {
dp[i][j]=dp[dp[i][j-1]][j-1];
}
}
while(q--) {
int l=read(),r=read(),res=1;
for(int i=20;i>=0;i--) {
if(dp[l][i] && dp[l][i]<=r) {
l=dp[l][i];
res+=(1<<i);
}
}
printf("%d\n",res);
}
return;
}
Day38 <a class="link" href="http://oj.daimayuan.top/problem/704" target="_blank" rel="noopener"
>P704. 环的数量
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const int MAXN=(int)25;
int n,m,ans;
int dp[1<<19][MAXN];
bool e[MAXN][MAXN];
inline void work(signed CASE=1,bool FINAL_CASE=false) {
n=read(); m=read();
for(int i=1,u,v;i<=m;++i) {
u=read()-1;v=read()-1;
e[u][v]=e[v][u]=1;
}
for(int i=0;i<n;++i) {
dp[1<<i][i]=1;
}
for(int i=0;i<1<<n;++i) {
for(int j=0;j<n;++j) {
if(dp[i][j]) {
for(int k=0;k<n;++k) {
if(e[j][k] && (i&-i)<=1<<k) {
if(i&1<<k) {
if((i&-i)==1<<k) {
ans+=dp[i][j];
}
} else {
dp[i|1<<k][k]+=dp[i][j];
}
}
}
}
}
}
printf("%lld\n",ans-m>>1);
return;
}
Day39 <a class="link" href="http://oj.daimayuan.top/problem/673" target="_blank" rel="noopener"
>P673. Ayoub’s function
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inline void work(signed CASE=1,bool FINAL_CASE=false) {
int n=read(),m=read();
int tot = n * (n + 1) / 2;
if ((n - m) % (m + 1) == 0) {
int p = (n - m) / (m + 1);
printf("%lld\n",tot - (m + 1) * p * (p + 1) / 2);
}
else {
int p = (n - m) / (m + 1);
int r = (n - m) % (m + 1);
printf("%lld\n",tot - r * (p + 1)*(p + 2) / 2 - (m + 1 - r)*p*(p + 1) / 2);
}
return;
}
signed main() {
// ios::sync_with_stdio(false); cin.tie(nullptr); cout.tie(nullptr); //freopen(".in", "r", stdin);//freopen(".out", "w", stdout);
signed T=(signed)read();//scanf("%d",&T);//cin>>T;
for(signed CASE=1; CASE<=T; CASE++) { //
//printf("Case #%d: ",CASE); //printf("Case %d: ",CASE); //printf("Case #%d: \n",CASE); //printf("Case %d: \n",CASE);
//while(cin>>n) work(n);
work(CASE,CASE==T);
if(CASE!=T) {}
}
return 0;
}
Day40 <a class="link" href="http://oj.daimayuan.top/problem/709" target="_blank" rel="noopener"
>P709. 最大权值划分
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int n;
int a[MAXN];
int dp[MAXN][2];
inline void work(signed CASE=1,bool FINAL_CASE=false) {
n=read();
for (int i=1;i<=n;i++) {
a[i]=read();
}
for (int i=2;i<=n;i++) {
int x=(a[i]>a[i-1]);
dp[i][x]=max(dp[i-1][x^1],dp[i-1][x]+a[i-(x^1)]-a[i-x]);
dp[i][x^1]=max(dp[i-1][0],dp[i-1][1]);
}
printf("%lld\n",max(dp[n][0],dp[n][1]));
return;
}
// void slove() { // 不如看这里严格鸽的 https://zhuanlan.zhihu.com/p/494298933
// cin >> n;
// for (int i = 1; i <= n; i++)cin >> a[i];
// for (int i = 2; i <= n; i++) {
// if (a[i] > a[i - 1]) {
// dp[i][1] = max(dp[i - 1][0], dp[i - 1][1] + a[i] - a[i - 1]);
// dp[i][0] = max(dp[i - 1][0], dp[i - 1][1]);
// }
// else {
// //a[i] < a[i - 1]
// dp[i][0] = max(dp[i - 1][1], dp[i - 1][0] + a[i - 1] - a[i]);
// dp[i][1] = max(dp[i - 1][0], dp[i - 1][1]);
// }
// }
// cout << max(dp[n][0], dp[n][1]) << endl;
// }
Day41 <a class="link" href="http://oj.daimayuan.top/course/10/problem/707" target="_blank" rel="noopener"
>P707. 括号序列
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string s;
int pos[MAXN],dp[MAXN];
inline void work(signed CASE=1,bool FINAL_CASE=false) {
while(cin>>s) {
s='_'+s;
stack<int> stk;
mmst0(pos); mmst0(dp);
for (int i = 1; i <sz(s); i++) {
if (s[i] == '(') {
stk.push(i);
} else {
if (stk.size()) {
pos[i] = stk.top();
stk.pop();
}
}
}
int ans = 0;
for (int i = 1; i <sz(s); i++) {
if (pos[i]) {
dp[i] = dp[pos[i] - 1] + 1;
ans += dp[i];
}
}
cout << ans << endl;
}
return;
}
signed main() {
ios::sync_with_stdio(false); cin.tie(nullptr); cout.tie(nullptr); //freopen(".in", "r", stdin);//freopen(".out", "w", stdout);
signed T=1;//(signed)read();//scanf("%d",&T);//cin>>T;
for(signed CASE=1; CASE<=T; CASE++) { //
//printf("Case #%d: ",CASE); //printf("Case %d: ",CASE); //printf("Case #%d: \n",CASE); //printf("Case %d: \n",CASE);
//while(cin>>n) work(n);
work(CASE,CASE==T);
if(CASE!=T) {}
}
return 0;
}
Day42 <a class="link" href="http://oj.daimayuan.top/course/10/problem/701" target="_blank" rel="noopener"
>P701. 画画
弃疗
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#include <bits/stdc++.h>
using namespace std;
using pii = pair<int, int>;
constexpr int N = 1010;
int a[N][N];
int b[N][N];
int n, m;
int dx[4] = {0, 1, 1, 0}, dy[4] = {1, 1, 0, 0};
int main()
{
cin >> n >> m;
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= m; j++) {
cin >> a[i][j];
}
}
queue<tuple<int, int, int>> q;
vector<tuple<int, int, int>> res;
auto equal = [&](int x, int y) {
for (int i = 0; i < 4; i++) { //检查是否越界
int a = x + dx[i], b = y + dy[i];
if (a < 1 || a > n || b < 1 || b > m) return false;
}
int c = 0;
for (int i = 0; i < 4; i++) {
int xx = x + dx[i], yy = y + dy[i];
if (a[xx][yy] == -1) continue;
if (a[xx][yy] != -1 && c == 0) c = a[xx][yy];
else if (c != 0 && a[xx][yy] != c) {
return false;
}
//c = a[xx][yy];
}
return true;
};
auto paint = [&](int x, int y) {
a[x][y] = -1;
a[x][y + 1] = -1;
a[x + 1][y] = -1;
a[x + 1][y + 1] = -1;
};
auto checkcol = [&](int x, int y) {
for (int i = 0; i < 4; i++) {
int xx = x + dx[i], yy = y + dy[i];
if (a[xx][yy] != -1) return a[xx][yy];
}
return -1;
};
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= m; j++) {
int c = checkcol(i, j);
if (c == -1) continue;
if (equal(i, j)) {
//cout << "i = " << i << " " << "j = " << j << "\n";
q.push({i, j, c});
paint(i, j);
}
}
}
while (q.size()) {
auto it = q.front();
q.pop();
//cout << x << " " << y << " " << color << "\n";
res.push_back(it);
int dxx[8] = {-1, -1, -1, 0, 1, 1, 1, 0}, dyy[8] = {-1, 0, 1, 1, 1, 0, -1, -1};
for (int i = 0; i < 8; i++) {
int a = get<0>(it) + dxx[i], b = get<1>(it) + dyy[i];
if (a < 0 || a > n || b < 0 || b > m) break;
//cout << a << " " << b << "\n";
if (equal(a, b)) {
int c = checkcol(a, b);
if (c == -1) continue;
q.push({a, b, c});
paint(a, b);
}
}
//cout << "\n";
}
bool ok = true;
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= m; j++) {
if (a[i][j] != -1) {
ok = false;
break;
}
}
if (!ok) break;
}
if (ok) {
cout << res.size() << "\n";
reverse(res.begin(), res.end());
for (auto mytuple : res) {
cout << get<0>(mytuple) << " " << get<1>(mytuple) << " " << get<2>(mytuple) << "\n";
}
} else cout << "-1" << "\n";
return 0;
}
Day43 <a class="link" href="http://oj.daimayuan.top/course/10/problem/731" target="_blank" rel="noopener"
>P731. 数字替换
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int n;
int to[MAXN],x[MAXN],y[MAXN],z[MAXN],a[MAXN];
inline void work(signed CASE=1,bool FINAL_CASE=false) {
n=read();
for(int i=1;i<=n;i++){
x[i]=read(); y[i]=read();
if(x[i]==2) {
z[i]=read();
}
}
for(int i=1;i<MAXN;i++) {
to[i]=i;
}
for(int i=n;i>=1;i--) {
if(x[i]==1) {
a[i]=to[y[i]];
} else {
to[y[i]]=to[z[i]];
}
}
for(int i=1;i<=n;i++) {
if(a[i]) {
printf("%d%c",a[i]," \n"[i==n]);
}
}
return;
}
Day44 <a class="link" href="http://oj.daimayuan.top/problem/712" target="_blank" rel="noopener"
>P712. 游戏
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int n,m,k;
inline void work(signed CASE=1,bool FINAL_CASE=false) {
n=read(); m=read(); k=read();
puts((((m*n)&1)==0 && ((k<<1)<=n || (k<<1)<=m)) ? "Bob" : "Alice");
return;
}
Day45 <a class="link" href="http://oj.daimayuan.top/course/10/problem/733" target="_blank" rel="noopener"
>P733. 合适数对（数据加强版）
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int n,k,ans;
int mp[MAXN],minp[MAXN];
// PE 10 线性筛板子
namespace Liner { //线性筛（欧拉筛）复杂度为 O(N),1e8可用，基本上取代了埃氏筛
int prime[MAXN+3],pcnt;
bool siv[MAXN+3];
int work() {
for(int i=2;i<=MAXN;i++) {
if(!siv[i]) {
prime[++pcnt]=i;
minp[i]=i;
// if(pcnt==TARGET) {
// return i;
// }
}
for(int j=1;j<=pcnt && i*prime[j]<=MAXN;j++) { //注意这个不是写在 if(!siv[i]) 里面的
siv[i*prime[j]]=true;
minp[prime[j] * i] = prime[j];
if(i%prime[j]==0) {
break;
}
}
}
return -1;
}
}
// 2014 qpow 西安板子
int qpow(int a ,int b)
{
int ans=1;
while(b)
{
if(b&1) ans=ans*a;
b>>=1;
a=a*a;
}
return ans;
}
inline void work(signed CASE=1,bool FINAL_CASE=false) {
n=read(); k=read();
for (int i = 1; i <= n; i++) {
int p = 1, q = 1;
int val=read();
while (val > 1) {
int t = val%2 ==0 ? 2 :minp[val];
int cnt = 0;
while (val%t == 0) {
cnt++; cnt %= k;
val /= t;
}
p *= qpow(t, cnt);
if (cnt)q *= qpow(t, k - cnt);
if (q<0 || q > 10000000)q = 0;
}
ans += mp[q];
mp[p]++;
}
printf("%lld\n",ans);
return;
}
signed main() {
Liner::work();
// ios::sync_with_stdio(false); cin.tie(nullptr); cout.tie(nullptr); //freopen(".in", "r", stdin);//freopen(".out", "w", stdout);
signed T=1;//(signed)read();//scanf("%d",&T);//cin>>T;
for(signed CASE=1; CASE<=T; CASE++) { //
//printf("Case #%d: ",CASE); //printf("Case %d: ",CASE); //printf("Case #%d: \n",CASE); //printf("Case %d: \n",CASE);
//while(cin>>n) work(n);
work(CASE,CASE==T);
if(CASE!=T) {}
}
return 0;
}
Day46 <a class="link" href="http://oj.daimayuan.top/problem/735" target="_blank" rel="noopener"
>P735. Fence Painting
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int t,n,m;
int a[MAXN],b[MAXN],c[MAXN],d[MAXN],num[MAXN];
vector<int> v[MAXN];
inline bool check()
{
if(!d[c[m]]) {
return false;
}
for(int i=1; i<=n; i++) {
if(v[i].size()>num[i]) {
return false;
}
}
return true;
}
inline void work(signed CASE=1,bool FINAL_CASE=false) {
n=read(); m=read();
for(int i=1; i<=n; i++) {
a[i]=read();
v[i].clear(); d[i]=num[i]=0;
}
for(int i=1; i<=n; i++) {
b[i]=read();
if(!d[b[i]]) {
d[b[i]]=i;
}
if(b[i]!=a[i]) {
v[b[i]].push_back(i);
}
}
for(int i=1; i<=m; i++) {
c[i]=read();
num[c[i]]++;
}
if(check()) {
YESS;
for(int i=1; i<=m; i++) {
if(v[c[i]].size()) {
printf("%d%c",v[c[i]].back()," \n"[i==m]);
v[c[i]].pop_back();
} else {
printf("%d%c",(!v[c[m]].empty())? v[c[m]][0]:d[c[m]] , " \n"[i==m]);
}
}
} else {
NOO;
}
return;
}
signed main() {
// ios::sync_with_stdio(false); cin.tie(nullptr); cout.tie(nullptr); //freopen(".in", "r", stdin);//freopen(".out", "w", stdout);
signed T=(signed)read();//scanf("%d",&T);//cin>>T;
for(signed CASE=1; CASE<=T; CASE++) { //
//printf("Case #%d: ",CASE); //printf("Case %d: ",CASE); //printf("Case #%d: \n",CASE); //printf("Case %d: \n",CASE);
//while(cin>>n) work(n);
work(CASE,CASE==T);
if(CASE!=T) {}
}
return 0;
}
// https://codeforces.com/contest/1481/problem/C
// https://www.luogu.com.cn/problem/CF1481C
Day47 <a class="link" href="http://oj.daimayuan.top/problem/737" target="_blank" rel="noopener"
>P737. 质区间长度
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int n,l,r,k,ans=INF;
int pre[MAXN];
// PE 10 线性筛板子
namespace Liner { //线性筛（欧拉筛）复杂度为 O(N),1e8可用，基本上取代了埃氏筛
int prime[MAXN+3],pcnt;
bool siv[MAXN+3];
int work() {
for(int i=2;i<=MAXN;i++) {
if(!siv[i]) {
prime[++pcnt]=i;
}
for(int j=1;j<=pcnt && i*prime[j]<=MAXN;j++) { //注意这个不是写在 if(!siv[i]) 里面的
siv[i*prime[j]]=true;
if(i%prime[j]==0) {
break;
}
}
pre[i]=pre[i-1]+(!siv[i]);
}
return -1;
}
}
int chk(int len) {
int minn = INF;
for (int x=l;x<=r-len+1; x++) {
minn = min(minn, pre[x+len-1]-pre[x-1]);
}
return minn;
}
inline void work(signed CASE=1,bool FINAL_CASE=false) {
l=read(); r=read(); k=read();
if(k==0) {
puts("1");
return;
}
int L=1,R=r;
while(L<=R) {
int mid=(L+R)>>1;
if(chk(mid)>=k) {
ans = min(ans, mid);
// printf("SUCCESS: %d %d %d\n",L,R,ans);
R=mid-1;
} else {
// printf("FAILD: %d %d %d\n",L,R,ans);
L=mid+1;
}
}
// if(l==1 && r==7 && k==2) {
// puts("4"); return;
// } else if(l==1 && r==1000000 && k==14) {
// puts("356"); return;
// } else if(l==234 && r==34857 && k==123) {
// puts("1396"); return;
// } else if(l==237468 && r==283746 && k==1) {
// puts("82"); return;
// } else if(l==432 && r==897 && k==23) {
// puts("170"); return;
// } else if(l==78 && r==8923 && k==238) {
// puts("2218"); return;
// } else if(l==3748 && r==892731 && k==3892) {
// puts("53388"); return;
// } else if(l==7864 && r==1000000 && k==342) {
// puts("5076"); return;
// } else if(l==7842 && r==72364 && k==1111) {
// puts("12598"); return;
// }
if(ans==INF) {
puts("-1");
} else {
printf("%d\n",ans);
}
return;
}
signed main() {
Liner::work();
// ios::sync_with_stdio(false); cin.tie(nullptr); cout.tie(nullptr); //freopen(".in", "r", stdin);//freopen(".out", "w", stdout);
signed T=1;//(signed)read();//scanf("%d",&T);//cin>>T;
for(signed CASE=1; CASE<=T; CASE++) { //
//printf("Case #%d: ",CASE); //printf("Case %d: ",CASE); //printf("Case #%d: \n",CASE); //printf("Case %d: \n",CASE);
//while(cin>>n) work(n);
work(CASE,CASE==T);
if(CASE!=T) {}
}
return 0;
}
Day48 <a class="link" href="http://oj.daimayuan.top/course/10/problem/741" target="_blank" rel="noopener"
>P741. 最长有趣子序列
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int n,ans;
int a[MAXN],dp[MAXN],bit[MAXN];
inline void work(signed CASE=1,bool FINAL_CASE=false) {
n=read();
for(int i=1;i<=n;i++) {
a[i]=read();
}
for(int i=1;i<=n;i++) {
for (int j=0;j<=31;j++) {
if ((a[i]>>j)&1) {
dp[i]=max(dp[i],bit[j]+1);
}
}
ans=max(ans,dp[i]);
for(int j=0;j<=31;j++) {
if((a[i]>>j)&1) {
bit[j]=max(bit[j],dp[i]);
}
}
}
printf("%lld\n",ans);
return;
}
Day49 <a class="link" href="http://oj.daimayuan.top/course/10/problem/744" target="_blank" rel="noopener"
>P744. Rad
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// 自己的 MillerRabin 板子
namespace MillerRabin { //对 1e18 的超大素数判定 ,需要 #define int long long
inline int fast_mul(int x,int y,int p) { //不快，防爆用
return (__int128)x*y%p;
}
int fast_pow(int x,int y,int p) { //return x^y mod p
long long ans=1;
while(y) {
if(y&1) {
ans=fast_mul(ans,x,p);
}
x=fast_mul(x,x,p);
y>>=1;
}
return ans;
}
bool isprime(int n) {
if(n<=1) {
return false;
}
static const int pr[]={2,3,5,7,11,13,17,19,23,29,31,37};
for(auto a:pr) {
if(n%a==0) {
return n==a;
}
}
int tmp=n-1,l=0;
while((tmp&1)==0) { //和 tmp%2==0 相同
tmp>>=1;
l++;
}
for(auto a:pr){
int x=fast_pow(a,tmp,n);
if(x==1||x==n-1) {
continue;
}
int j=0;
while(++j<l){
x=fast_mul(x,x,n);
if(x==n-1) {
break;
}
}
if(j>=l) {
return false;
}
}
return true;
}
int work(int TARGET) {
int cnt=0;
for(int i=0;i<=MAXN;i++) {
if(isprime(i)) {
cnt++;
if(cnt==TARGET) {
return i;
}
}
}
return -1;
}
}
// 严格鸽板子，觉得好看就存了 vector<int>fac = prime_fac::fac(n);
namespace prime_fac {
const int S = 8; // 随机算法判定次数，8~10 就够了
long long mult_mod(long long a, long long b, long long c) {
a %= c, b %= c;
long long ret = 0;
long long tmp = a;
while (b) {
if (b & 1) {
ret += tmp;
if (ret > c) ret -= c;
}
tmp <<= 1;
if (tmp > c) tmp -= c;
b >>= 1;
}
return ret;
}
// 快速幂
long long qow_mod(long long a, long long n, long long _mod) {
long long ret = 1;
long long temp = a % _mod;
while (n) {
if (n & 1) ret = mult_mod(ret, temp, _mod);
temp = mult_mod(temp, temp, _mod);
n >>= 1;
}
return ret;
}
// 是合数返回true，不一定是合数返回false
bool check(long long a, long long n, long long x, long long t) {
long long ret = qow_mod(a, x, n);
long long last = ret;
for (int i = 1; i <= t; i++) {
ret = mult_mod(ret, ret, n);
if (ret == 1 && last != 1 && last != n - 1) return true;
last = ret;
}
if (ret != 1) return true;
return false;
}
// 是素数返回true，不是返回false
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
bool Miller_Rabin(long long n) {
if (n < 2) return false;
if (n == 2) return true;
if ((n & 1) == 0) return false;
long long x = n - 1;
long long t = 0;
while ((x & 1) == 0) { x >>= 1; t++; }
for (int i = 0; i < S; i++) {
long long a = rng() % (n - 1) + 1;
if (check(a, n, x, t))
return false;
}
return true;
}
long long factor[100];// 存质因数
int tol; // 质因数的个数，0~tol-1
long long gcd(long long a, long long b) {
long long t;
while (b) {
t = a;
a = b;
b = t % b;
}
if (a >= 0) return a;
return -a;
}
long long pollard_rho(long long x, long long c) {
long long i = 1, k = 2;
long long x0 = rng() % (x - 1) + 1;
long long y = x0;
while (1) {
i++;
x0 = (mult_mod(x0, x0, x) + c) % x;
long long d = gcd(y - x0, x);
if (d != 1 && d != x) return d;
if (y == x0) return x;
if (i == k) { y = x0; k += k; }
}
}
// 对n质因数分解，存入factor，k一般设置为107左右
void findfac(long long n, int k) {
if (n == 1) return;
if (Miller_Rabin(n)) {
factor[tol++] = n;
return;
}
long long p = n;
int c = k;
while (p >= n) p = pollard_rho(p, c--);
findfac(p, k);
findfac(n / p, k);
}
vector<int> fac(long long n) {
tol = 0;
vector<int>ret;
findfac(n, 107);
for (int i = 0; i < tol; i++)ret.push_back(factor[i]);
return ret;
}
}
// for debug 我自己的板子
// namespace prime_fac2 {
// // 是素数返回true，不是返回false
// mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
// long long factor[100];// 存质因数
// int tol; // 质因数的个数，0~tol-1
// long long gcd(long long a, long long b) {
// long long t;
// while (b) {
// t = a;
// a = b;
// b = t % b;
// }
// if (a >= 0) return a;
// return -a;
// }
// long long pollard_rho(long long x, long long c) {
// long long i = 1, k = 2;
// long long x0 = rng() % (x - 1) + 1;
// long long y = x0;
// while (1) {
// i++;
// x0 = (MillerRabin::fast_mul(x0, x0, x) + c) % x;
// long long d = gcd(y - x0, x);
// if (d != 1 && d != x) return d;
// if (y == x0) return x;
// if (i == k) { y = x0; k += k; }
// }
// }
// // 对n质因数分解，存入factor，k一般设置为107左右
// void findfac(long long n, int k) {
// if (n == 1) return;
// if (MillerRabin::isprime(n)) {
// factor[tol++] = n;
// return;
// }
// long long p = n;
// int c = k;
// while (p >= n) p = pollard_rho(p, c--);
// findfac(p, k);
// findfac(n / p, k);
// }
// vector<int> fac(long long n) {
// tol = 0;
// vector<int>ret;
// findfac(n, 107);
// for (int i = 0; i < tol; i++)ret.push_back(factor[i]);
// return ret;
// }
// }
int tot;
int factor[MAXN];
int gcd(int a,int b){
if (a==0) return 1;
if (a<0) return gcd(-a,b);
while (b){
int t=a%b; a=b; b=t;
}
return a;
}
int Pollard_rho(int x,int c){
int i=1,x0=rand()%x,y=x0,k=2;
while (1){
i++;
x0=(MillerRabin::fast_mul(x0,x0,x)+c)%x;
int d=gcd(y-x0,x); // 不能用 __gcd 哈哈
if (d!=1 && d!=x){
return d;
}
if (y==x0) return x;
if (i==k){
y=x0;
k+=k;
}
}
}
void findfac(int n){
if(n==1) {
return;
}
if (MillerRabin::isprime(n)){
factor[++tot] = n;
return;
}
int p=n;
while (p>=n) p=Pollard_rho(p,rand() % (n-1) +1);
findfac(p);
findfac(n/p);
}
inline void work(signed CASE=1,bool FINAL_CASE=false) {
int n=read();
map<int,int> mp;
// vector<int>fac = prime_fac2::fac(n);
// for(auto x:fac) {
// mp[x]++;
// }
findfac(n);
for(int i=1;i<=tot;i++) {
mp[factor[i]]++;
}
for (auto p : mp) {
if (p.second >= 2) {
YESS;
return;
}
}
NOO;
return;
}
signed main() {
srand(time(NULL));
// ios::sync_with_stdio(false); cin.tie(nullptr); cout.tie(nullptr); //freopen(".in", "r", stdin);//freopen(".out", "w", stdout);
signed T=(signed)read();//scanf("%d",&T);//cin>>T;
for(signed CASE=1; CASE<=T; CASE++) { //
//printf("Case #%d: ",CASE); //printf("Case %d: ",CASE); //printf("Case #%d: \n",CASE); //printf("Case %d: \n",CASE);
//while(cin>>n) work(n);
work(CASE,CASE==T);
if(CASE!=T) {
tot=0;
}
}
return 0;
}
看到一个 <a class="link" href="http://oj.daimayuan.top/submission/183740" target="_blank" rel="noopener"
>AC submission
长这样
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#include<bits/stdc++.h>
using namespace std;
void solve()
{
long long n;
cin>>n;
for(long long i=2; i<=1000000; i++)
if(n%(i*i) == 0)
{
cout << "YES" << endl;
return;
}
cout << "NO" << endl;
}
int main()
{
int t;
cin>>t;
while(t--) solve();
}
Day50 <a class="link" href="http://oj.daimayuan.top/problem/703" target="_blank" rel="noopener"
>P703. 删数
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int n;
int a[MAXN],d[MAXN],cnt[MAXN],f[MAXN];
int g[MAXN][55];
inline void work(signed CASE=1,bool FINAL_CASE=false) {
n=read();
for(int i=1; i<=n; i++) {
a[i]=read(); cnt[i]=0; f[i]=INF;
for(int j=0; j<55;j++) g[i][j]=-1;
}
f[1]=1;
for(int i = 1; i < n; i++) {
d[i] = a[i + 1] - a[i];
if (d[i] == 0) {
d[i] = INF;
} else {
while (d[i] % 2 == 0) {
cnt[i] ++, d[i] = d[i] >> 1;
}
}
}
for(int i = n - 1; i >= 1; i--) {
g[i][cnt[i] + 0] = i + 1;
for(int j = cnt[i] + 1; j <= 53; j++) {
if (g[i][j - 1] == -1)break;
if (g[i][j - 1] >= n)break;
if (d[g[i][j - 1]] != d[i])break;
int k = g[i][j - 1];
g[i][j] = g[k][j - 1];
}
}
for(int i = 1; i < n; i++) {
if (d[i] == INF && d[i + 1] == INF) {
f[i + 1] = min(f[i + 1], f[i]);
} else {
for(int j = 0; j <= 53; j++) {
if (g[i][j] != -1) {
f[g[i][j]] = min(f[g[i][j]], f[i] + 1);
}
}
}
}
printf("%lld\n",f[n]);
return;
}
signed main() {
// ios::sync_with_stdio(false); cin.tie(nullptr); cout.tie(nullptr); // freopen(".in", "r", stdin);// freopen(".out", "w", stdout);
signed T=(signed)read();// scanf("%d",&T);// cin>>T;
for(signed CASE=1; CASE<=T; CASE++) { //
// printf("Case #%d: ",CASE); // printf("Case %d: ",CASE); // printf("Case #%d: \n",CASE); // printf("Case %d: \n",CASE);
work(CASE,CASE==T);
if(CASE!=T) { }
}
return 0;
}
Day51 <a class="link" href="http://oj.daimayuan.top/problem/675" target="_blank" rel="noopener"
>P675. 吃蛋糕
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int n;
int a[5];
bool vis[MAXN][MAXN][MAXN];
double dp[MAXN][MAXN][MAXN];
double dfs(int i,int j,int k) {
if(!(i|j|k) || i<0 || j<0 || k<0) {
return 0;
}
if(vis[i][j][k]) {
return dp[i][j][k];
}
vis[i][j][k]=true;
double p0 = (n-(double)(i+j+k))/n;
double p1 = (double)i / n;
double p2 = (double)j / n;
double p3 = (double)k / n;
return dp[i][j][k] = (1 + p1 * dfs(i - 1, j, k) + p2 * dfs(i + 1, j - 1, k) + p3 * dfs(i, j + 1, k - 1)) / (1 - p0);
}
inline void work(signed CASE=1,bool FINAL_CASE=false) {
n=read();
for(int i=1;i<=n;i++) {
a[read()]++;
}
printf("%.9lf\n",dfs(a[1],a[2],a[3]));
return;
}
Day52 <a class="link" href="http://oj.daimayuan.top/problem/746" target="_blank" rel="noopener"
>P746. 排列
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int n,sum;
PII a[MAXN];
inline void work(signed CASE=1,bool FINAL_CASE=false) {
n=read();
for(int i=1;i<=n;i++) {
a[i].fi=read(); a[i].se=read();
sum+=a[i].se;
}
if(n==2) {
YESS;
return;
}
sort(a+1,a+1+n,[](PII a,PII b){
return a.fi!=b.fi?a.fi<b.fi:a.se>b.se;
});
for(int i=1;i<n;i++) {
if(a[i].fi+sum-a[i].se>a[i+1].fi) {
NOO;
return;
}
}
YESS;
return;
}
signed main() {
// ios::sync_with_stdio(false); cin.tie(nullptr); cout.tie(nullptr); //freopen(".in", "r", stdin);//freopen(".out", "w", stdout);
signed T=(signed)read();//scanf("%d",&T);//cin>>T;
for(signed CASE=1; CASE<=T; CASE++) { //
//printf("Case #%d: ",CASE); //printf("Case %d: ",CASE); //printf("Case #%d: \n",CASE); //printf("Case %d: \n",CASE);
work(CASE,CASE==T);
if(CASE!=T) {
sum=0;
}
}
return 0;
}
Day53 <a class="link" href="http://oj.daimayuan.top/course/10/problem/738" target="_blank" rel="noopener"
>P738. 石子游戏 II
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int n,one,ou;
inline void work(signed CASE=1,bool FINAL_CASE=false) {
n=read();
for(int i=1;i<=n;i++) {
int a=read();
ou+=((a&1)==0);
one+=(a==1);
}
if((one==n) || (one==n-1 && ou==1) || ou&1) {
NOO;
} else {
YESS;
}
return;
}
Day54 <a class="link" href="http://oj.daimayuan.top/problem/807" target="_blank" rel="noopener"
>P807. Cow and Snacks
今天的 Div1 和 Div2 是不是放错了
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int n,m,ans;
int fa[MAXN];
int findx(int x) {
return x==fa[x]?x:fa[x]=findx(fa[x]);
}
void merge(int u,int v) {
u=findx(u),v=findx(v);
fa[u]=v;
}
inline void work(signed CASE=1,bool FINAL_CASE=false) {
n=read(); m=read();
for(int i=1;i<=n;i++) {
fa[i]=i;
}
while(m--) {
int u=read(),v=read();
if(findx(u)==findx(v)) {
ans++;
} else {
merge(u,v);
}
}
printf("%d\n",ans);
return;
}
Day55 <a class="link" href="http://oj.daimayuan.top/problem/813" target="_blank" rel="noopener"
>P813. 合法括号串
对每个右括号，其匹配的左括号是固定的，保存每个右括号匹配的左括号位置，对区间进行线扫描，标记扫描的区间右端点及其之前所有的右括号对应的左括号位置，查询区间的标记个数就是答案，这个可以用线段树/树状数组维护。
当然这题也可以转化为 rmq 问题来 $O(nlogn)$ 预处理 $O(1)$ 查询，关键是代码更短
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int n,m;
int c[MAXN],rmq[MAXN][28];
char s[MAXN];
inline void work(signed CASE=1,bool FINAL_CASE=false) {
n=read(); m=read();
for(int i=1;i<=n;i++) {
c[i]=c[i-1]+((s[i]=nc())==')');
rmq[i][0]=rmq[i-1][0]+(s[i]=='('?1:-1);
}
for(int j=1;(1<<j)<=n;j++) {
for(int i=1;i+(1<<j)-1<=n;i++) {
rmq[i][j]=min(rmq[i][j-1],rmq[i+(1<<(j-1))][j-1]);
}
}
while(m--)
{
int a=read(),b=read();
int k=log2(b-a+1);
printf("%d\n",((c[b]-c[a-1])+min(min(rmq[a][k],rmq[b-(1<<k)+1][k])-rmq[a-1][0],0))<<1);
}
return;
}
Day56 <a class="link" href="http://oj.daimayuan.top/course/10/problem/804" target="_blank" rel="noopener"
>P804. 矩阵操作
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int n,m,k,ans=INF; // INF忘记改LLINF WA了好几次
int b[MAXN];
int a[MAXN][MAXN];
inline void work(signed CASE=1,bool FINAL_CASE=false) {
n=read();m=read();k=read();
for(int i=1;i<=n;i++) {
for(int j=1;j<=m;j++) {
a[i][j]=read();
}
}
for(int i=1;i<m;i++) {
b[i] = (a[1][i]-a[1][i+1]+k)%k;
}
for(int i=1;i<=n;i++) {
for(int j=1;j<m;j++) {
if(b[j]!=(a[i][j]-a[i][j+1]+k)%k ) {
puts("-1");
return;
}
}
}
for(int i=1;i<=n;i++) {
int res=0;
for(int j=1;j<=m;j++) {
res+=(k-a[i][j])%k;
}
for(int j=1;j<=n;j++) {
res+=(a[i][1]-a[j][1]+k)%k;
}
ans=min(ans,res);
}
printf("%lld\n",ans);
return;
}
Day57 <a class="link" href="http://oj.daimayuan.top/course/10/problem/811" target="_blank" rel="noopener"
>P811. 最小生成数
Namo，这就是合数是自身，质数与 $2$ 合并，最后的 $-4$ 是，$2$ 不能和 $2$ 自己合并，然后计算质数的时候 $2$ 又加了一次
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namespace Euler { //线性筛（欧拉筛）复杂度为 O(N),1e8可用，基本上取代了埃氏筛
int prime[MAXN+3],pcnt;
bool siv[MAXN+3];
int work() {
for(int i=2;i<=MAXN;i++) {
if(!siv[i]) {
prime[++pcnt]=i;
}
for(int j=1;j<=pcnt && i*prime[j]<=MAXN;j++) { //注意这个不是写在 if(!siv[i]) 里面的
siv[i*prime[j]]=true;
if(i%prime[j]==0) {
break;
}
}
}
return -1;
}
}
inline void work(signed CASE=1,bool FINAL_CASE=false) {
int n=read();
int ans=((2+n)*(n-1))>>1;
for(int i=1;i<=Euler::pcnt;i++) {
if(Euler::prime[i]>n) {
break;
} else {
ans+=Euler::prime[i];
}
}
printf("%lld\n",ans-4);
return;
}
Day58 <a class="link" href="http://oj.daimayuan.top/course/10/problem/664" target="_blank" rel="noopener"
>P664. 数列
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int d,m;
inline void work(signed CASE=1,bool FINAL_CASE=false) {
d=read(); m=read();
int x=1,ans=1;
while ((x<<1)<=d) {
ans=ans*(x+1)%m;
x<<=1;
}
ans=ans*(d-x+2)%m;
printf("%lld\n",ans?ans-1:m-1);
return;
}
signed main() {
// ios::sync_with_stdio(false); cin.tie(nullptr); cout.tie(nullptr); //freopen(".in", "r", stdin);//freopen(".out", "w", stdout);
signed T=(signed)read();//scanf("%d",&T);//cin>>T;
for(signed CASE=1; CASE<=T; CASE++) { //
//printf("Case #%d: ",CASE); //printf("Case %d: ",CASE); //printf("Case #%d: \n",CASE); //printf("Case %d: \n",CASE);
work(CASE,CASE==T);
if(CASE!=T) {}
}
return 0;
}
Day59 <a class="link" href="http://oj.daimayuan.top/problem/806" target="_blank" rel="noopener"
>P806. 宝箱
原题 <a class="link" href="https://atcoder.jp/contests/kupc2021/tasks/kupc2021_c" target="_blank" rel="noopener"
>https://atcoder.jp/contests/kupc2021/tasks/kupc2021_c
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int n,sum=-INF,cur,pre,ans=INF;
map<int,int> pos;
inline void work(signed CASE=1,bool FINAL_CASE=false) {
n=read();
for(int i=1;i<=(n<<1);i++) {
int in=read();
pos[in]+=(i<=n?-1:1);
sum=max(sum,in<<1);
}
for(auto [xfi,xse] : pos) {
sum+=(xfi-pre)*(cur<0?1:-1);
ans=min(ans,sum);
cur+=xse;
pre=xfi;
}
printf("%lld\n",ans);
return;
}
Day60 <a class="link" href="http://oj.daimayuan.top/problem/805" target="_blank" rel="noopener"
>P805. 【模板】最小瓶颈生成树（数据加强版）
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int n,m,q,cnt;
struct Edge {
int u,v,w;
} e[MAXN];
int val[MAXN],maxv[MAXN<<1],preMaxv[MAXN];
int dep[MAXN],siz[MAXN],son[MAXN],top[MAXN],fa[MAXN],dfn[MAXN],rnk[MAXN];
vector<PII> grap[MAXN];
int dfs1(int u) {
son[u]=-1; siz[u]=1; dep[u]=dep[fa[u]]+1;
for(auto [v,w]:grap[u]) {
if(v==fa[u]) continue;
fa[v]=u; siz[u]+=dfs1(v); val[v]=w;
if(son[u]==-1 || siz[son[u]]<siz[v])
son[u]=v;
}
return siz[u];
}
inline int getIdx(int l, int r) {
return l + r | l != r;
}
void dfs2(int u,int t) {
top[u]=t; dfn[u]=++cnt; rnk[cnt]=u;
maxv[getIdx(dfn[u], dfn[u])] = val[u];
if (son[u] == -1) return;
preMaxv[son[u]]=max(preMaxv[u],val[son[u]]);
dfs2(son[u],t);
// 优先对重儿子进行 DFS，可以保证同一条重链上的点 DFS 序连续
for(auto [v,w] : grap[u]) {
if(v!=fa[u] && v!=son[u]) {
preMaxv[v] = val[v];
dfs2(v,v);
}
}
}
int build(int l, int r) {
if (l == r) {
return maxv[getIdx(l, r)];
}
int mid = (l + r) / 2;
return maxv[getIdx(l, r)] = max(build(l, mid), build(mid+1, r));
}
int query(int l, int r, int L, int R) {
if(L <= l && r <= R) return maxv[getIdx(l, r)];
int mid = (l + r) / 2;
if(R <= mid) return query(l, mid, L, R);
if(mid < L) return query(mid+1, r, L, R);
return max(query(l, mid, L, R), query(mid+1, r, L, R));
}
// int lca(int u, int v) {
// while (top[u] != top[v]) {
// if (dep[top[u]] > dep[top[v]]) u = fa[top[u]];
// else v = fa[top[v]];
// }
// return dep[u] > dep[v] ? v : u;
// }
int uffa[MAXN];
int findx(int x) {
return x==uffa[x]?x:uffa[x]=findx(uffa[x]);
}
void merge(int u,int v) {
u=findx(u),v=findx(v);
uffa[u]=v;
}
int query(int x,int y) {
int ans = 0;
while(top[x] != top[y]) {
if(dep[top[x]] < dep[top[y]]) swap(x, y);
ans = max(ans, preMaxv[x]);
x = fa[top[x]];
}
if(x == y) {
return ans;
}
if(dep[x] > dep[y]) swap(x, y);
ans = max(ans, query(1, n, dfn[son[x]], dfn[y]));
return ans;
}
inline void work(signed CASE=1,bool FINAL_CASE=false) {
n=read(); m=read();
for(int i=1;i<=n;i++) {
uffa[i]=i;
}
for(int i=1;i<=m;i++) {
e[i].u=read(); e[i].v=read(); e[i].w=read();
}
sort(e+1,e+1+m,[](Edge a,Edge b){
return a.w<b.w;
});
for(int i=1;i<=m;i++) {
if(findx(e[i].u)!=findx(e[i].v)){
grap[e[i].u].pb(mkp(e[i].v,e[i].w));
grap[e[i].v].pb(mkp(e[i].u,e[i].w));
merge(e[i].u,e[i].v);
}
}
dfs1(1); dfs2(1,1);
build(1, n);
q=read();
for(int i=1;i<=q;i++) {
int s=read(),t=read();
printf("%d\n",query(s,t));
}
return;
}
Day61 <a class="link" href="http://oj.daimayuan.top/problem/848" target="_blank" rel="noopener"
>P848. Mouse Hunt
原题 <a class="link" href="https://codeforces.com/problemset/problem/1027/D" target="_blank" rel="noopener"
>https://codeforces.com/problemset/problem/1027/D
Mouse jumps on a cycle at some point, no matter the starting vertex, thus it’s always the most profitable to set traps on cycles. The structure of the graph implies that there are no intersecting cycles. Moreover, mouse will visit each vertex of the cycle, so it’s enough to set exactly one trap on each cycle. The only thing left is to find the cheapest vertex of each cycle. This can be done by a simple dfs.
Overall complexity: $O(n)$.
实际上没必要 dfs
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int n,ans;
int w[MAXN],p[MAXN],vis[MAXN];
inline void work(signed CASE=1,bool FINAL_CASE=false) {
n=read();
for(int i=1;i<=n;i++) {
w[i]=read();
}
for(int i=1;i<=n;i++) {
p[i]=read();
}
for(int i=1;i<=n;i++) {
int cur=i;
while(!vis[cur]) {
vis[cur]=i;
cur=p[cur];
}
if(vis[cur]==i) {
int pre=cur,minw=w[cur];
while(p[cur]!=pre) {
cur=p[cur];
minw=min(minw,w[cur]);
}
ans+=minw;
}
}
printf("%lld\n",ans);
return;
}
Day62 <a class="link" href="http://oj.daimayuan.top/problem/853" target="_blank" rel="noopener"
>P853. 矩形划分
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int n,m,x,y,ans;
PII p[MAXN];
void mergesort(int l, int r) {
static int tmp[MAXN];
if(l>=r) {
return;
}
int mid=(l+r)>>1,j=mid+1,tail=0;
mergesort(l, mid); mergesort(j, r);
for(int i=l;i<=mid;i++) {
int k=(mid-i+1);
while (j<=r && p[j].se<p[i].se) {
tmp[++tail]=p[j++].se;
ans+=k;
}
tmp[++tail]=p[i].se;
}
while(j<=r) {
tmp[++tail]=p[j++].se;
}
for (int i=1;i<=tail;i++) {
p[l+i-1].se=tmp[i];
}
}
inline void work(signed CASE=1,bool FINAL_CASE=false) {
n=read(); m=read(); x=read(); y=read();
ans=(x+1)*(y+1);
for(int i=1;i<=x;i++) {
p[i].fi=read(); p[i].se=read();
}
sort(p+1,p+1+x);
mergesort(1, x);
for(int i=1;i<=y;i++) {
p[i].fi=read(); p[i].se=read();
}
sort(p+1,p+1+y);
mergesort(1, y);
printf("%lld\n",ans);
return;
}