Codeforces 894.A QAQ

“QAQ” is a word to denote an expression of crying. Imagine “Q” as eyes with tears and “A” as a mouth.

Now Diamond has given Bort a string consisting of only uppercase English letters of length n. There is a great number of “QAQ” in the string (Diamond is so cute!).

Bort wants to know how many subsequences “QAQ” are in the string Diamond has given. Note that the letters “QAQ” don’t have to be consecutive, but the order of letters should be exact.

Input
The only line contains a string of length n (1 ≤ n ≤ 100). It’s guaranteed that the string only contains uppercase English letters.

Output
Print a single integer — the number of subsequences “QAQ” in the string.

Examples
input
QAQAQYSYIOIWIN
output
4
input
QAQQQZZYNOIWIN
output
3
Note
In the first example there are 4 subsequences “QAQ”: “QAQAQYSYIOIWIN”, “QAQAQYSYIOIWIN”, “QAQAQYSYIOIWIN”, “QAQAQYSYIOIWIN”.

题意：

给一个长度 不超过 100 的字符串 ， 统计 QAQ 的出现次数 ，（QAQ的出现次数 ：QAQ 不一定要是连续的 ， 但是位置上 一定是 A的两边各有一个Q）

题解：

以A为轴的话，两边的Q可以任意组合，故组合方法一共（A左边Q的数量*A右边Q的数量）种

CODE

#include <stdio.h>
#include <iostream>
#include <string.h>
using namespace std;
int main()
{
    int num[200];
    char ch[200];
    gets(ch);
    int l=strlen(ch);
    int n=0;
    memset(num,0,sizeof(num));
    for(int i=0;i<l;i++)
    {
        if(ch[i]=='Q')
            n++;
        num[i]=n;
    }
    int sum=0;
    for(int i=0;i<l;i++)
    {
        if(ch[i]=='A')
            sum+=num[i]*(num[l-1]-num[i]);
    }
    cout << sum << endl;
    return 0;
}