codeforce_894A_字符串水_找出子串所有QAQ可以不连续-优快云博客

本文介绍了一种算法，用于计算字符串中不连续子序列'QAQ'的数量。该算法通过遍历字符串并记录'Q'和'A'字符的位置来工作，最终计算出所有可能的'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
4
input
QAQQQZZYNOIWIN
output
3
Note
In the first example there are 4 subsequences “QAQ”: “QAQAQYSYIOIWIN”, “QAQAQYSYIOIWIN”, “QAQAQYSYIOIWIN”, “QAQAQYSYIOIWIN”.

找出子串所有QAQ可以不连续

#include<bits/stdc++.h>
#define ll long long
using namespace std;
int main()
{
    char s[1111];
    while(cin>>s)
    {
        ll i,j,len=strlen(s),q[111],a[111],lq=0,la=0;
        for(i=0; i<len; i++)
        {
            if(s[i]=='Q')
                q[lq++]=i;
            else if(s[i]=='A') a[la++]=i;
        }
        ll ans=0;
       // cout<<lq<< " "<<la<<endl;
        for(i=0; i<la; i++)
        {
            ll tt=0;
            for(j=0; j<lq; j++)
                if(q[j]<a[i])
                    tt++;
                else break;
            ans+=tt*(lq-tt);
        }
        cout<<ans<<endl;
    }
    return 0;
}