PAT甲级 1093 Count PAT‘s (25 分)

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该程序读取一个包含P、A、T字符的字符串，并计算其中包含'P'开头，中间是'A'，以'T'结尾的PAT子串的数目。通过维护左右P和T的计数，对每个'A'字符计算其左侧的P数量和右侧的T数量，累加到答案上。最后输出结果模1000000007。

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>

using namespace std;
const int Mod = 1000000007;
int leftP[100005];

int main() {
    string s;
    int rightT = 0;
    int ans = 0;
    cin >> s;
    for (int i = 0; i < s.size(); ++i) {
        if (i > 0)
            leftP[i] = leftP[i - 1];
        if (s[i] == 'P')
            leftP[i]++;
    }
    for (int i = s.size() - 1; i >= 0; --i) {
        if (s[i] == 'T')
            rightT++;
        else if (s[i] == 'A')
            ans = (ans + rightT * leftP[i]) % Mod;
    }
    cout << ans << endl;
    return 0;
}