HDU 5056 Boring count_a-good string description: you are given a string -优快云博客

本文链接：https://blog.youkuaiyun.com/qq_35846109/article/details/79405914

本文介绍了一种算法，用于解决特定条件下字符串中限定子串的数量计算问题。该算法使用了滑动窗口的方法，并提供了完整的代码实现。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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

   6 15 21 
 

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

Output

For each case, output a line contains the answer.

Sample Input

    3 abc 1 abcabc 1 abcabc 2 
  

Sample Output

    6 15 21 
  

尺取法，有一点要注意：关于字符串子串数目的计算问题

int cnt = 0;
    for(int i = 0; i < len; i++)
    {
        cnt = i + 1;
    }

代码：

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<char, int>P;
const int inf = 0x3f3f3f3f;
char s[100005];
int cnt[30];
int main()
{
    int t, k;
    scanf("%d", &t);
    while(t--)
    {
        memset(cnt, 0, sizeof(cnt));
        scanf("%s", &s);
        scanf("%d", &k);
        int len = strlen(s);
        ll ans = 0;
        int it, is;
        for(it = 0, is = 0; it < len; it++)
        {
            cnt[s[it]-'a']++;
            if(cnt[s[it]-'a']>k)
            {
                for(;is <= it; is++)
                {
                    cnt[s[is]-'a']--;
                    if(cnt[s[it]-'a']<=k)
                        break;
                }
                is++;
            }
            ans += (it-is+1);
        }
        printf("%lld\n", ans);
    }
}