1040 有几个PAT (25分)

#include<stdio.h>
#include<string>
#include<iostream>
using namespace std;
int main(){
    string str;
    getline(cin,str);
    int len=str.size();
    int leftnump[len]={0},rightnumt[len]={0};
    int ans=0;
    if(str[0]=='P')leftnump[0]=1;
    for(int i=1;i<len;i++){
        if(str[i]=='P')leftnump[i]=leftnump[i-1]+1;
        else leftnump[i]=leftnump[i-1];

    }
    if(str[len-1]=='T')rightnumt[len-1]=1;
    for(int i=len-2;i>=0;i--){
        if(str[i]=='T')rightnumt[i]=rightnumt[i+1]+1;
        else rightnumt[i]=rightnumt[i+1];
        if(str[i]=='A') ans=(ans+rightnumt[i]*leftnump[i])%1000000007;
    }
    printf("%d\n",ans);
    return 0;
}

第一次，参考算法笔记完成，30min