• Codeforce -An impassioned circulation of affection


    尺取

    复杂度 (n)

     1 # include <cstdio>
     2 # include <iostream>
     3 # include <algorithm>
     4 # include <cstring>
     5 using namespace std;
     6 
     7 const int maxn=1500+5;
     8 int n,m;
     9 char s[maxn];
    10 int d;
    11 char ss[3];
    12 
    13 int main(){
    14   while(scanf("%d",&n)!=EOF){
    15     scanf("%s",s);
    16     scanf("%d",&m);
    17     while(m--){
    18         scanf("%d%s",&d,ss);
    19         int res=0,ls=0,rs=0,sum=0;
    20         for(ls=0;ls<n;ls++){
    21           while(rs<n&&sum+(s[rs]!=ss[0])<=d) {
    22             sum+=(s[rs]!=ss[0]);
    23             rs++;
    24           }
    25           res=max(res,rs-ls);
    26           sum-=(s[ls]!=ss[0]);
    27         }
    28     printf("%d
    ",res);
    29     }
    30   }
    31   return 0;
    32 }
    View Code
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  • 原文地址:https://www.cnblogs.com/lintanxi/p/6970476.html
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