HDU 5056 Boring count（不超过k个字符的子串个数）

Boring count

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

Total Submission(s): 828    Accepted Submission(s): 342

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

 

Source

field=problem&key=BestCoder+Round+%2311+%28Div.+2%29&source=1&searchmode=source" style="color:rgb(26,92,200); text-decoration:none">BestCoder Round #11 (Div. 2)

 

题目大意：给出一个字符串，求出子串中每个字母出现次数不超过k的个数。

解题思路：枚举字符串下标i，每次计算以i为结尾的符合条件的最长串。那么以i为结尾的符合条件子串个数就是最长串的长度。求和就可以。

代码例如以下：

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <string>
#include <vector>
#include <deque>
#include <list>
#include <set>
#include <map>
#include <stack>
#include <queue>
#include <numeric>
#include <iomanip>
#include <bitset>
#include <sstream>
#include <fstream>
#include <limits.h>
#define debug "output for debug\n"
#define pi (acos(-1.0))
#define eps (1e-6)
#define inf (1<<28)
#define sqr(x) (x) * (x)
#define mod 1000000007
using namespace std;
typedef long long ll;
typedef unsigned long long ULL;
#define maxn 100010
int n,m,t,k;
char str[maxn];
int num[maxn];
int main()
{
    scanf("%d",&t);
    while(t--)
    {
        memset(num,0,sizeof(num));
        scanf("%s",str);
        scanf("%d",&k);
        int k1=strlen(str);
        int pos=0;
        ll ans=0;
        for(int i=0; i<k1; i++)
        {
            num[str[i]-'a']++;
            if(num[str[i]-'a']>k)
            {
                while(str[pos]!=str[i])
                {
                    num[str[pos]-'a']--;
                    pos++;
                }
                num[str[pos]-'a']--;
                pos++;
            }
            ans+=(i-pos+1);
        }
        printf("%I64d\n",ans);
    }
    return 0;
}