Codeforces 894A(QAQ)

最新推荐文章于 2024-02-22 13:57:15 发布

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最新推荐文章于 2024-02-22 13:57:15 发布

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分类专栏：字符串处理 OJ刷题

本文链接：https://blog.youkuaiyun.com/qq_40922859/article/details/81175497

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这篇博客讨论了如何计算一个给定的仅包含大写英文字母的字符串中，"QAQ"子序列出现的次数。题目描述了一个简单的算法问题，要求找出不连续但顺序精确的"QAQ"序列的数量。给出的例子中，字符串中存在4个"QAQ"子序列。

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"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

Input

QAQQQZZYNOIWIN

Output

Note

In the first example there are 4 subsequences "QAQ": "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN".

#include<bits/stdc++.h>
typedef long long ll;
using namespace std;
int main(){
	string a;
	while(cin>>a){
	  ll ans=0;
	  ll cnt=0;
	  for(ll i=0;i<a.length();i++){
	  	  if(a[i]=='Q') cnt++;
	  }
	  ll temp=0;
	  for(ll i=0;i<a.length();i++){
	  	 if(a[i]=='Q'){
	  	   temp++;	
		 }
		 else if(a[i]=='A'){
		 	ans+=temp*(cnt-temp);
		 }
	  }
	  cout<<ans<<endl;
	}
	return 0;
}