Boring count

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

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

思路：O（n）的算法，枚举i，求出以每个i结尾的满足条件的串的个数，即以i结尾的最长串的长度。

源代码：

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>

#define N 111111
using namespace std;

typedef long long LL;

char str[100001];
int cnt[30];
int k;

int main()
{
    int T;
    cin>>T;
    while(T--)
    {
        cin>>(str+1)>>k;
        memset(cnt,0,sizeof cnt);
        int n=strlen(str+1);
        int pos=0;
        LL res=0;
        for(int i=1;i<=n;i++)
        {
            cnt[str[i]-'a']++;
            if(cnt[str[i]-'a']>k)
            {
                pos++;
                while(str[pos]!=str[i])
                {
                    cnt[str[pos]-'a']--;
                    pos++;
                }
                cnt[str[pos]-'a']--;
            }
            res+=i-pos;
        }
        printf("%I64d\n",res);
    }
    return 0;
}