CodeForces - 894A-QAQ（思维）

最新推荐文章于 2020-01-30 20:54:03 发布

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最新推荐文章于 2020-01-30 20:54:03 发布

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文章标签：数据结构与算法

原文链接：http://www.cnblogs.com/Staceyacm/p/10781880.html

本文介绍了一种算法，用于计算给定字符串中不连续但精确顺序的QAQ子序列数量。通过两次遍历字符串，分别计算每个'A'前后的'Q'的数量，然后将这些数量相乘，得到所有可能的QAQ组合。

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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!).

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

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".

题解：去求字符串里面有多少个QAQ字串，可以不连续

代码：

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>

using namespace std;

int main()
{
	
	string str;
	cin>>str;
	int s2,s1,s=0;
	for(int t=0;t<str.length();t++)
	{
		
		if(str[t]=='A')
		{
			s1=0;
			s2=0;
			for(int j=0;j<t;j++)
			{
				if(str[j]=='Q')
				s1++;
			}
			for(int j=t+1;j<str.length();j++)
			{
				if(str[j]=='Q')
				s2++;
			}
			s+=s1*s2;
		}
	}
	cout<<s<<endl;
	
	return 0;
}

转载于:https://www.cnblogs.com/Staceyacm/p/10781880.html