PAT 1093. Count PAT's

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本文介绍了一种高效算法，用于计算给定字符串中特定子串PAT出现的次数。通过从后向前遍历字符串并统计不同阶段的字符数量，最终得出PAT子串的总数。为避免巨大数值，计算过程中取1000000007的余数。

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:

2

思路：从后往前依次统计T的个数，AT的个数和PAT的个数

#include <stdio.h>
#include <string.h>
char ch[100010];
int main()
{
	while(scanf("%s", &ch) != EOF)
	{
		int length = strlen(ch);
		int cntT = 0, cntAT = 0, cntPAT = 0;
		for(int i = length - 1; i >= 0; i--)
		{
			if(ch[i] == 'T')
			{
				cntT++;
			}
			if(ch[i] == 'A')
			{
				cntAT += cntT;
				cntAT %= 1000000007;
			}
			if(ch[i] == 'P')
			{
				cntPAT += cntAT;
				cntPAT %= 1000000007;
			}
		}
		printf("%d\n", cntPAT);
	}
	return 0;
}