2023 Hubei Provincial Collegiate Programming Contest
Posted sakuya726
tags:
篇首语：本文由小常识网(cha138.com)小编为大家整理，主要介绍了2023 Hubei Provincial Collegiate Programming Contest相关的知识，希望对你有一定的参考价值。
记录一下和队友VP的这场湖北省赛，有空会加上赛后自己的复盘分析和补题情况。
C.Darkness I
#include<bits/stdc++.h> using namespace std; #define rg register #define maxn 105050 #define int long long inline int read() int x=0,f=1; char c=getchar(); while(c<\'0\'||c>\'9\') if(c==\'-\') f=-1; c=getchar(); while(c>=\'0\'&&c<=\'9\') x=(x<<3)+(x<<1)+c-48; c=getchar(); return x*f; int n,m; signed main() n=read(); m=read(); cout<<min(n,m)+ceil((max(n,m)-min(n,m))/2.0);
F. Inverse Manacher
#include<bits/stdc++.h> using namespace std; #define rg register #define maxn 2850500 #define mod 998244353 #define int long long inline int read() int x=0,f=1; char c=getchar(); while(c<\'0\'||c>\'9\') if(c==\'-\') f=-1; c=getchar(); while(c>=\'0\'&&c<=\'9\') x=(x<<3)+(x<<1)+c-48; c=getchar(); return x*f; int n; int a[maxn]; char ans[maxn]; int max_r; signed main() n=read(); for(rg int i=0;i<2*n+2;++i) a[i]=read(); ans[2]=\'a\'; max_r=2; for(rg int i=3;i<2*n+2;++i) if(i%2==0) for(rg int j=max_r+2;j<=i-1+a[i];j+=2) ans[j]=ans[2*i-j]; max_r=j; // cout<<"i="<<i<<" maxr="<<max_r<<" rr="<<i-1+a[i]<<endl; if(2*i-max_r-2>0&&max_r+2<=i+1+a[i]) max_r+=2; ans[max_r]=((ans[2*i-max_r]==\'a\')?\'b\':\'a\'); else for(rg int j=max_r+2;j<=i-1+a[i];j+=2) ans[j]=ans[2*i-j]; max_r=j; // cout<<"i="<<i<<" maxr="<<max_r<<" rr="<<i-1+a[i]<<endl; if(2*i-max_r-2>0&&max_r+2<=i+1+a[i]) max_r+=2; ans[max_r]=((ans[2*i-max_r]==\'a\')?\'b\':\'a\'); for(rg int i=2;i<2*n+2;i+=2) cout<<ans[i];
H. Binary Craziness
#include<bits/stdc++.h> using namespace std; #define rg register #define maxn 1850500 #define mod 998244353 #define int long long inline int read() int x=0,f=1; char c=getchar(); while(c<\'0\'||c>\'9\') if(c==\'-\') f=-1; c=getchar(); while(c>=\'0\'&&c<=\'9\') x=(x<<3)+(x<<1)+c-48; c=getchar(); return x*f; int n,m; int dg[maxn]; int num[maxn],temp[maxn]; bool flag[maxn]; struct node int val,tot; ; stack<node> q; inline int f(int a,int b) return (a&b)*(a|b)*(a^b)%mod; int ans; signed main() n=read(); m=read(); for(rg int i=1;i<=m;++i) int a,b; a=read(); b=read(); dg[a]++; dg[b]++; for(rg int i=1;i<=n;++i) num[dg[i]]++; for(rg int i=1;i<=n;++i) if(flag[dg[i]]==0) flag[dg[i]]=1; node a; a.val=dg[i]; a.tot=num[dg[i]]; q.push(a); temp[q.size()]=a.val; int cnt=q.size(); for(rg int i=1;i<=cnt;++i) int val=temp[i]; int tot=(num[val]+1)*num[val]/2; ans+=f(val,val)*tot; ans%=mod; for(rg int j=i+1;j<=cnt;++j) ans+=((num[temp[i]]*num[temp[j]])*f(val,temp[j])); ans%=mod; cout<<ans;
J. Expansion
#include<bits/stdc++.h> using namespace std; #define rg register #define maxn 550500 #define mod 998244353 #define int long long inline int read() int x=0,f=1; char c=getchar(); while(c<\'0\'||c>\'9\') if(c==\'-\') f=-1; c=getchar(); while(c>=\'0\'&&c<=\'9\') x=(x<<3)+(x<<1)+c-48; c=getchar(); return x*f; int n; int a[maxn]; int t; int s[maxn],max_top; inline int find(int now) for(rg int i=now+1;i<=n;++i) if(s[i]>=max_top&&s[i+1]<s[i]) return i; return n; inline int get(int l,int r) int minn=99999999999; int sum=0; for(rg int i=l+1;i<=r;++i) sum+=s[i]; minn=min(minn,sum); max_top=max(max_top,s[i]); return minn; int sum,st; signed main() n=read(); for(rg int i=1;i<=n;++i) a[i]=read(),s[i]=s[i-1]+a[i]; for(rg int i=1;i<=n;++i) sum+=s[i]; if(s[i+1]<=s[i]&&s[i]>0) st=i; max_top=s[i]; break; if(s[i]<0) cout<<-1; return 0; if(s[n]<0) cout<<-1; return 0; for(rg int i=st;i<n;++i) int nxt=find(i); int nd=get(i,nxt); // cout<<"i="<<i<<" sum="<<sum<<" nxt="<<nxt<<" nd="<<nd<<endl; if(sum>=abs(nd)) for(rg int j=i+1;j<=nxt;++j) sum+=s[j]; i=nxt-1; continue; else // cout<<"i="<<i<<" sum="<<sum<<" nd="<<nd<<" nxt="<<nxt<<" s[i]="<<s[i]<<" t+="<<ceil((double)(1.0*(abs(sum+nd)))/(double)(1.0*s[i]))<<" t="<<t<<endl; t+=ceil((double)(1.0*(abs(sum+nd)))/(double)(1.0*s[i])); sum=ceil((double)(1.0*(abs(sum+nd)))/(double)(1.0*s[i]))*s[i]+sum; for(rg int j=i+1;j<=nxt;++j) sum+=s[j]; i=nxt-1; // cout<<"t="<<t<<endl; cout<<t+n;
K. Dice Game
#include<bits/stdc++.h> using namespace std; #define rg register #define maxn 105050 #define int long long #define mod 998244353 inline int read() int x=0,f=1; char c=getchar(); while(c<\'0\'||c>\'9\') if(c==\'-\') f=-1; c=getchar(); while(c>=\'0\'&&c<=\'9\') x=(x<<3)+(x<<1)+c-48; c=getchar(); return x*f; inline long long qp(long long a,long long b) long long ans=1; while(b>0) if(b&1) ans=(ans*a)%mod; a=(long long)(a*a)%mod; b>>=1; return ans%mod; int n,m; //p/q p*q^(998244351) mod 998244353 signed main() n=read(); m=read(); // cout<<qp(64,998244351)*27%mod; for(rg int i=1;i<=m;++i)//[1,m]---x [1,x-1] [x+1,m] (x-1)/m的概率赢 (m-x)/m的概率输 cout<<qp((int)qp(m-1,n),998244351)*(int)qp(m-i,n)%mod<<" ";
M. Different Billing
#include<bits/stdc++.h> using namespace std; #define rg register #define maxn 105050 #define int long long inline int read() int x=0,f=1; char c=getchar(); while(c<\'0\'||c>\'9\') if(c==\'-\') f=-1; c=getchar(); while(c>=\'0\'&&c<=\'9\') x=(x<<3)+(x<<1)+c-48; c=getchar(); return x*f; int x,y; //1000 2500 signed main() x=read(); y=read(); for(rg int i=0;i<=x;++i) long long sumb=i*1000; if((y-sumb)==0) cout<<x-i<<" "<<i<<" "<<0; return 0; else if((y-sumb)>0&&(y-sumb)%2500==0) cout<<max((int)0,x-i-(y-sumb)/2500)<<" "<<i<<" "<<(y-sumb)/2500; return 0; cout<<-1; return 0;
HZNU Training 23 for Zhejiang Provincial Competition 2020
拓扑排序，从上往下回溯，挺好奇差分约束为啥就不对。拓扑排序有明显的优先级。
#include<bits/stdc++.h> using namespace std; #define pb push_back typedef long long ll; const int inf=0x3f3f3f3f; typedef long long ll; typedef double db; const int N=1e4+50; vector<int>e[N]; int n,m; int in[N],a[N]; int topo(){ int cnt=0; queue<int>Q; for(int i=1;i<=n;i++)if(in[i]==0)cnt++,Q.push(i); while(!Q.empty()){ int u=Q.front();Q.pop(); for(int i=0;i<e[u].size();i++){ int v=e[u][i];--in[v]; if(!in[v]){ Q.push(v),cnt++; a[v]=max(a[v],a[u]+1); } } } if(cnt<n)return -1; int sum=0; for(int i=1;i<=n;i++)sum+=888+a[i]; return sum; } int main(){ while(~scanf("%d %d",&n,&m)){ memset(in,0,sizeof in); memset(a,0,sizeof a); for(int i=1;i<=n;i++)e[i].clear(); for(int i=1,u,v;i<=m;i++){ scanf("%d %d",&u,&v); e[v].pb(u);in[u]++; } cout<<topo()<<endl; } // system("pause"); return 0; }
G - G
带权并查集
以上是关于2023 Hubei Provincial Collegiate Programming Contest的主要内容，如果未能解决你的问题，请参考以下文章
2023 CCPC Henan Provincial Collegiate Programming Contest
Zhejiang Provincial Programming Contest 2007 Evolution (矩阵快速幂)
HZNU Training 23 for Zhejiang Provincial Competition 2020
The 10th Shandong Provincial Collegiate Programming Contest（8/13）
The 2018 ACM-ICPC China JiangSu Provincial Programming Contest(第六场)