Boring count
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 228 Accepted Submission(s): 90
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.
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.
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
abc
1
abcabc
1
abcabc
2
Sample Output
6
15
21
15
21
Source
之前用一种动态规划式方法怎么用怎么TLE ，看了一下解题报告，发现有O(n)算法。觉得很是强大，........哎，还是太弱哇...╮(╯▽╰)╭
代码：
#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
const int maxn=;
char str[maxn];
int sac[];
int main()
{
int cas,st,pos,k;
scanf("%d",&cas);
while(cas--)
{
memset(sac,,sizeof(sac));
scanf("%s %d",str,&k);
__int64 ans=;
st=;
for(int i=;str[i]!='\0';i++)
{
pos=str[i]-'a';
sac[pos]++;
if(sac[pos]>k)
{
while(str[st]!=str[i])
{
sac[str[st]-'a']--;
st++;
}
sac[str[st]-'a']--;
st++;
}
ans+=(i-st+);
}
printf("%I64d\n",ans);
}
return ;
}