Count PAT's (25)

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题目描述
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.

输入描述:
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.

输出描述:
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.

输入例子:
APPAPT

输出例子:
2

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大意：
给出一个只包含PAT三个字符的字符串，求按照字符的出现顺序能够找出多少个不必相邻的字符组合组成“PAT”。答案需要对 1000000007求余。

分析
很简单的一道题目，不难看出求法。直接写代码即可。

#include<iostream>
#include<string>
using namespace std;
const int maxn = 100009;
const int mod = 1000000007;

struct Node{	//message of char 'A'
	int lp;	//number of 'P' in left 
	int rt;	//number of 'T' in right
}nodea[maxn];

int ans;
int main(){
	//receive input data
	string input;
	int tlp=0,trt = 0;
	cin >> input;
	//recorde lp of each char 'A';
	for (int i = 0; i < input.length(); i++){
		if (input[i] == 'P') tlp++;
		else if (input[i] == 'A') nodea[i].lp = tlp;
	}
	//recorde rt of each char 'A'
	for (int i = input.length() - 1; i >= 0; i--){
		if (input[i] == 'T') trt++;
		else if (input[i] == 'A')nodea[i].rt = trt;
	}
	//caculate the answer
	for (int i = 0; i < maxn; i++){
		int add = nodea[i].lp*nodea[i].rt;
		if (add!=0){
			ans += add;
			ans %= mod;
		}
	}
	cout << ans << endl;
}