hdu 5056(尺取法)

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

 

题目大意：就是问在子串中每个小写字母出现次数不超过k次的个数

解题思路：尺取法，维护i和j使得str[i]到str[j]之间的所有小写字母的个数不超过k。

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;

const int maxn = 100005;
int k,len,cnt[200];
char str[maxn];

bool check()
{
	for(char i = 'a'; i <= 'z'; i++)
		if(cnt[i] > k) return false;
	return true;
}

int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		memset(cnt,0,sizeof(cnt));
		cin >> str >> k;
		len = strlen(str);
		__int64 ans = 0;
		for(int i = 0, j = 0; i < len; i++)
		{
			cnt[str[i]]++;
			while(check() == false) 
			{
				cnt[str[j]]--;
				j++;
			}
			ans += i - j + 1;
		}
		printf("%I64d\n",ans);
	}
	return 0;
}