noi.ac#228 book

分析

代码

#include<bits/stdc++.h>
using namespace std;
#define int long long
const int inf =1e18;
char s[1000100];
int a[1000100],pre[1000100],sur[1000100],mn[1000100];
signed main(){
    int n,m,p,q,i,j,k,x,y,Ans=inf;
    cin>>n>>p>>q>>x>>y;
    scanf("%s",s+1);
    for(i=1;i<=n;i++)a[i]=(s[i]=='-'?-1:1);
    for(i=1;i<=n;i++)pre[i]=pre[i-1]+a[i];
    for(i=1;i<=n;i++)mn[i]=min(mn[i-1],pre[i]);
    k=(q-p-pre[n])/2;
    for(i=1;i<=n;i++){
      int sum=min(pre[n]-pre[n-i+1]+mn[n-i+1],sur[i-1]);
      int t=max(0ll,(-p-sum+1)/2);
      Ans=min(Ans,(abs(k-t)+t)*x+(i-1)*y);
      sur[i]=min(sur[i-1]+a[n-i+1],0ll);
    }
    cout<<Ans;
    return 0;
}

 

posted @ 2019-09-30 13:10  水题收割者  阅读(169)  评论(0编辑  收藏  举报