1093 count PAT's

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本文介绍了一个算法问题，即如何计算给定字符串中特定子串'PAT'出现的次数。输入为一个仅包含字符P、A、T的大字符串，输出为这些特定子串的数量，结果需对1000000007取模。

The string APPAPT contains two PAT's as substrings. The first one is formed by the 2nd, the 4th, and the 6th characters, and the second one is formed by the 3rd, the 4th, and the 6th characters.

Now given any string, you are supposed to tell the number of PAT's contained in the string.

Input Specification:

Each input file contains one test case. For each case, there is only one line giving a string of no more than 10⁵ characters containing only P, A, or T.

Output Specification:

For each test case, print in one line the number of PAT's contained in the string. Since the result may be a huge number, you only have to output the result moded by 1000000007.

Sample Input:

APPAPT

Sample Output:

#include<iostream>
#include<string>
#include<vector>
using namespace std;
int main(){
	for(string str;cin>>str;){
		int all_t = 0;
		for(int i = 0;i < str.length();i++){
			if(str[i] == 'T'){
				all_t ++;
			}
		}
		int sum = 0;
		int count_P = 0;
		int count_T = 0;
		for(int i = 0;i < str.length();i++){
			if(str[i] == 'P'){
				count_P++;
			}else if(str[i] == 'T'){
				count_T++;
			}else{
				sum+=(count_P*(all_t-count_T))%1000000007;
				sum = sum % 1000000007;
			}
		}
		cout<<sum<<endl;
	}
	return 0;
}