Codeforce -An impassioned circulation of affection

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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 }
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原文地址:https://www.cnblogs.com/lintanxi/p/6970476.html