codeforces 894A QAQ

计数字符串中QAQ子序列

最新推荐文章于 2024-06-24 08:26:42 发布

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本文介绍了一种方法来计算给定字符串中特定子序列QAQ出现的次数。通过遍历字符串并检查所有可能的子序列组合来实现，确保字母顺序正确且不连续。

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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!).

$\text{[math]}$ 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 
    
      Copy
    
QAQAQYSYIOIWIN

     output 
    
      Copy
    
4

     input 
    
      Copy
    
QAQQQZZYNOIWIN

     output 
    
      Copy
    
3

Note

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

因为n的取值比较小，直接暴力

#include<bits/stdc++.h>  
using namespace std;  
typedef long long ll;  
const int inf = 0x3f3f3f3f;  
int main()   
{  
    // freopen("shuju.txt","r",stdin);
    string a;
    cin>>a;
    int len=a.length();
    int ans=0;
    for(int i=0;i<len-2;i++)
    {
        for(int j=i+1;j<len-1;j++)
        {
            for(int k=j+1;k<len;k++)
            {
                if(a[i]=='Q'&&a[j]=='A'&&a[k]=='Q')
                    ans++;
            }
        }
    }
    cout<<ans<<endl;
    return 0;  
}