A-LCT

带权并查集，维护一下每个点在当前树的深度和以它为根能找到的最深的深度。‘

#include<bits/stdc++.h>using namespace std;
typedef long long ll;
const int N = 1e6 + 100;int fa[N],ans[N],val[N];int find(int x){if(fa[x]==x)return x;int t=fa[x];fa[x]=find(fa[x]);val[x]+=val[t];return fa[x];
}
void merge(int x,int y){int xx=x;int yy=find(y);if(xx!=yy){fa[xx]=yy;val[xx]=val[y]+1;ans[yy]=max(ans[yy],val[xx]+ans[xx]);}
}void solve(){int n;cin>>n;for(int i=1;i<=n;i++){fa[i]=i;ans[i]=0;val[i]=0;}for(int i=1;i<n;i++){int u,v,c;cin>>u>>v>>c;merge(v,u);cout<<ans[c]<<" ";}cout<<'\n';}signed main() {ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);int T = 1;cin>>T;while (T--)solve();return 0;
}

C-Sort4

考虑环的长度分类讨论即可

#include <bits/stdc++.h>
using namespace std;
#define int long long
const int N=1e6+5;
int n,m,n1,sum;
int q,f[N],a[N];
void solve(){n1=0,sum=0;cin>>n;for(int i=1;i<=n;i++) cin>>a[i];for(int i=1;i<=n;i++){if(f[i]) continue;int x=i,q=0;while(!f[x]){f[x]=1;x=a[x];q++;}if(q>4){sum+=(q-1)/3;if((q-1)%3) sum++;if(q%3==2) n1++;}else if(q>1){sum++;if(q==2) n1++;}}cout<<sum-n1/2<<'\n';for(int i=1;i<=n;i++) f[i]=0;
} 
signed main(){ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);int _=1;cin>>_;while(_--)solve();return 0;
}

F-Good Tree

对于u和v，在最优的情况下如果把树构造成一条链，容易发现最左或右的端点以及中心点一定是最优的，考虑二分去找这个点，最后统计答案的时候手磨一下发现对于x是奇数的情况，只能是奇数节点。

#include<bits/stdc++.h>using namespace std;
#define int long long
typedef long long ll;
const int N = 1e6 + 10;
int qu(int x) {if(x&1) {return (x-1)*(x-1)/4;} else {return x*x/4-x/2;}
}
void solve() {int l,r,x;cin>>x;l=1,r=2e9+5;int mid,a,ans=0;while(l<=r) {mid=(l+r)>>1;a=qu(mid);if(a>x) r=mid-1;else if(a<x) l=mid+1;else {ans=mid;break;}}if(ans) cout<<ans<<'\n';else{if(l%2==0){if(qu(l)%2==x%2) cout<<l<<'\n';else cout<<l+1<<'\n';}else{cout<<l<<'\n';}}
}signed main() {ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);int T = 1;cin>>T;while (T--)solve();return 0;
}

G-Horse Drinks Water

#include <bits/stdc++.h>
using namespace std;
#define int long long
void solve(){int x1,y1,x2,y2;cin>>x1>>y1>>x2>>y2;int xx1=x1+x2,yy1=abs(y1-y2);int xx2=abs(x1-x2),yy2=y1+y2;double ans0=sqrt(xx1*xx1+yy1*yy1);double ans1=sqrt(xx2*xx2+yy2*yy2);cout << fixed << setprecision(20)<<min(ans0,ans1)<<'\n'; 
} 
signed main(){ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);int _=1;cin>>_;while(_--)solve();return 0;
}

#include<bits/stdc++.h>using namespace std;
#define int long long
typedef long long ll;
const int N = 1e6 + 10;int a[N];
void solve(){int n;cin>>n;int mmin=1e17;for(int i=1;i<=n;i++){cin>>a[i];mmin=min(mmin,a[i]);}if(n==1){cout<<0<<'\n';return;}int ans=0;for(int i=1;i<=n;i++){if(a[i]==mmin)continue;int k=a[i]-mmin;if(ans==0)ans=k;else ans=__gcd(k,ans);}cout<<ans<<'\n';
}signed main() {ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);int T = 1;cin>>T;while (T--)solve();return 0;
}

#include<bits/stdc++.h>using namespace std;
#define int long long
typedef long long ll;
const int N = 1e6 + 10;int d[N];
struct node{int x,y;
}a[N];
bool cmp(node xx,node yy){if(xx.x==yy.x)return xx.y<yy.y;else return xx.x<yy.x;
}
void solve(){int n,m;cin>>n>>m;for(int i=1;i<=m;i++){int u,v;cin>>u>>v;if(u>v)swap(u,v);a[i].x=u;a[i].y=v;}sort(a+1,a+m+1,cmp);for(int i=1;i<=m;i++){d[a[i].y]++;}int l=1;int p=1;int ans=0;int ll=0,rr=0;for(int i=2;i<=n;i++){if(d[i]==i-l)continue;else{ans+=(i-l)*(i-1-l)/2;if(rr>l&&rr<=i){ans-=(1+rr-l)*(rr-l)/2;}rr=i-1;ll=l;while(d[i]!=i-l&&l<=i){while(p<=m&&a[p].x==l){d[a[p].y]--;p++;}l++;}}}if(l!=n){ans+=(1+n-l)*(n-l)/2;if(rr>l&&rr<=n){ans-=(1+rr-l)*(rr-l)/2;}}cout<<ans+n<<'\n';
}signed main() {ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);int T = 1;
//    cin>>T;while (T--)solve();return 0;
}

J-Zero

dp i j表示前i位，j次幂的时候，比i-1位增加的期望

为什么说是增加呢。

假设一个值一开始是，当它变成，增加的其实可以用二项式定理计算。

这样通过逐步递推就可以推出前n位的k次方

#include<bits/stdc++.h>
using namespace std;
#define int long long
typedef long long ll;
const int N = 2e5 + 100;
const int mod=998244353;int ksm(int a,int b)
{int anss=1;while(b){if(b&1)anss=anss*a%mod;a=a*a%mod;b>>=1;}return anss;
}int P1[N],P2[N];
void init(int tot) {P1[0]=1;for(int i=1; i<=tot; i++)P1[i]=1ll*P1[i-1]*i%mod;P2[tot]=ksm(P1[tot],mod-2);for(int i=tot-1; i>=0; i--)P2[i]=1ll*P2[i+1]*(i+1)%mod;
}
ll C(ll n,ll m) {return 1ll*P1[n]*P2[m]%mod*P2[n-m]%mod;
}int dp[N][50];
void solve(){int n,k;cin>>n>>k;string s;cin>>s;s="+"+s;for(int i=1;i<=n;i++){if(s[i]=='0')continue;for(int j=0;j<=k;j++){int res=1;for(int l=0;l<=j;l++){res=(res+dp[i-1][l]*C(j,l)%mod)%mod;}if(s[i]=='?')res=res*ksm(2,mod-2)%mod;dp[i][j]=res;}}int ans=0;for(int i=1;i<=n;i++){ans+=dp[i][k];ans%=mod;}cout<<ans<<'\n';
}signed main() {init(40);ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);int T = 1;
//    cin>>T;while (T--)solve();return 0;
}