CF894A QAQ

CF894A QAQ

题意翻译

题目大意： 给定一个字符串，字符串长度<=100，现在要求出字符串中'QAQ'的个数，注意，'QAQ'可以不连续，只要有顺序就可以了，例如QABQ也有一个QAQ

Translated by @Hjl1777

题目描述

"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 nn . There is a great number of "QAQ" in the string (Diamond is so cute!).

illustration by 猫屋 https://twitter.com/nekoyaliuBort 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.

输入输出格式

输入格式：

 

The only line contains a string of length nn ( 1<=n<=1001<=n<=100 ). It's guaranteed that the string only contains uppercase English letters.

 

输出格式：

 

Print a single integer — the number of subsequences "QAQ" in the string.

 

输入输出样例

输入样例#1： 复制

QAQAQYSYIOIWIN

输出样例#1： 复制

4

输入样例#2： 复制

QAQQQZZYNOIWIN

输出样例#2： 复制

3

说明

In the first example there are 44 subsequences "QAQ": "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN".

 

思路：组合数学搞一下就可以了。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
string s;
int ans,t,q[100];
int main(){
    cin>>s;int len=s.size();
    for(int i=0;i<s.size();++i)    q[i]=s[i]=='Q'?++t:t;
    for(int i=0;i<s.size();++i)    if(s[i]=='A')    ans+=q[i]*(t-q[i]);
    cout<<ans;
}