hdu 5907 Find Q

Find Q
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 262144/131072 K (Java/Others)
Total Submission(s): 756 Accepted Submission(s): 369
Problem Description
Byteasar is addicted to the English letter 'q'. Now he comes across a string S consisting of lowercase English letters.
He wants to find all the continous substrings of S, which only contain the letter 'q'. But this string is really really long, so could you please write a program to help him?
Input
The first line of the input contains an integer T(1≤T≤10), denoting the number of test cases.
In each test case, there is a string S, it is guaranteed that S only contains lowercase letters and the length of S is no more than 100000.
Output
For each test case, print a line with an integer, denoting the number of continous substrings of S, which only contain the letter 'q'.
Sample Input
2
qoder
quailtyqqq
Sample Output
1

7

题意：对于每组数据，输出一行一个整数，即仅包含字母'q'的连续子串的个数。

<span style="font-family:Arial;">#include <iostream>
#include<stdio.h>
#include<string.h>
using namespace std;
char s[100005];
long long int f(int x){//连续的q算他的连续的子序列 其实用公式更加简单
      long long int r=0;
      for(int i=1;i<=x;i++){
          r+=i;
      }
      return r;
}
int main()
{
    long long int T,cot,sum,l;
    cin>>T;
    while(T--){
         scanf("%s",s);
         cot=0,sum=0,l=strlen(s);
         for(int i=0;i<=l;i++){
            if(s[i]=='q'){
                cot++;
            }else{
                if(cot)
                sum+=f(cot);
                cot=0;
            }
         }
         cout<<sum<<endl;
    }
    return 0;
}
</span>