hdu 5056 Boring count（尺取法）_尺取法 hdu-CSDN博客

本文链接：https://blog.csdn.net/acm_cxq/article/details/51858378

本文介绍了一个算法问题的解决方案，该问题要求计算字符串中所有子串的数量，这些子串中每个字母出现的次数不超过给定值K。通过使用滑动窗口技术和哈希映射的方法，有效地解决了这一问题。

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Boring count

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 928 Accepted Submission(s): 381

Problem Description

You are given a string S consisting of lowercase letters, and your task is counting the number of substring that the number of each lowercase letter in the substring is no more than K.

Input

In the first line there is an integer T , indicates the number of test cases.
For each case, the first line contains a string which only consist of lowercase letters. The second line contains an integer K.

[Technical Specification]
1<=T<= 100
1 <= the length of S <= 100000
1 <= K <= 100000

Output

For each case, output a line contains the answer.

Sample Input

  
  
   
   3
abc
1
abcabc
1
abcabc
2

Sample Output

题意：出现字母不能超过k次的串的个数

思路：尺取法推进，map hash即可。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <map>
using namespace std;
#define N 100050
char s[N],a[N];
map<char,int>mp;
int main()
{
    int T,t;
    scanf("%d",&T);
    while(T--)
    {
        mp.clear();
        scanf("%s %d",s,&t);
        int r=0,cnt=0;
        long long ans=0;
        int len=strlen(s);
        for(int i=0; i<len; i++)
        {
            if(mp[s[r]]<=t)
                ans+=r-i;
            else ans+=r-i-1;
            while(r<len&&mp[s[r]]<t)
            {
                mp[s[r]]++;
                ans++;
                r++;
            }
            mp[s[i]]--;
        }
        printf("%lld\n",ans);
    }
    return 0;
}