# Find Q

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/131072 K (Java/Others)
Total Submission(s): 263    Accepted Submission(s): 148

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

AC代码：

#include <iostream>
#include<algorithm>
#include<string.h>
#include<stdio.h>
using namespace std;
char c[110000];
__int64 getsum(__int64 n)   //等差数列前n项和
{
if(n%2==0)return n/2*(1+n);
else return (1+n)/2*n;
}
int main()
{
int n;
scanf("%d%*c",&n);
while(n--)
{
__int64 sum=0,s=0;
gets(c);
for(__int64 i=0; i<(__int64)strlen(c); i++)
{
if(c[i]=='q')s++;
else
{
if(s)sum+=getsum(s);
s=0;
}
}
printf("%I64d\n",sum+getsum(s));
}
return 0;
}

### 4 只已被捕捉

• CAISIDUO Chrome | 54.0.2810.2 Windows 7

这是什么鬼 完全看不懂啊！( ´╥ω╥`)

• 千千 Chrome | 55.0.2853.0 Windows 10

你猜~

• CAISIDUO Chrome | 54.0.2810.2 Windows 7

一片看不懂的鹰文和格式有点像js的代码？

• 千千 Mozilla FireFox | 50.0 Windows 10

是C++啦