codeforces 894 A题 QAQ -mystical_curve

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本文介绍了一个简单的算法问题：如何在一个大写的英文字符组成的字符串中找出所有不连续但顺序正确的“QAQ”子序列的数量。通过使用前缀和的方法来统计Q字符出现的次数，并结合A字符的位置来计算所有可能的QAQ组合。

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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.

Example
Input
QAQAQYSYIOIWIN
Output
4
Input
QAQQQZZYNOIWIN
Output
3
Note
In the first example there are 4 subsequences “QAQ”: “QAQAQYSYIOIWIN”, “QAQAQYSYIOIWIN”, “QAQAQYSYIOIWIN”, “QAQAQYSYIOIWIN”.

//水题
#include <iostream>
#include <cstring>
using namespace std;
int main()
{
    string s;int m[120];
    memset(m,0,sizeof(m));
    getline(cin,s);int k=0;
    for(int i=0;i<s.size();i++)
    {
        m[i]=k;
        if(s[i]=='Q')
            k++;
        //cout<<m[i]<<"  ";
    }
    m[s.size()]=k;
    //cout<<endl;
    int ans=0;
    for(int i=0;i<s.size();i++)
    {
        if(s[i]=='A'){
            //cout<<i<<endl;
            ans+=m[i]*(m[s.size()]-m[i]);
            //cout<<m[i]<<"  "<<m[s.size()]<<endl;
        }
    }
    cout<<ans<<endl;
    return 0;
}