Pat(A) 1093. Count PAT's (25)

原题目：

原题链接：https://www.patest.cn/contests/pat-a-practise/1093

1093. Count PAT’s (25)

---

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

题目大意

给定一个只包含P、A、T的字符串，求出子串为“PAT”的个数。

解题报告

利用排列组合的方法，对于任意一个A，假如其前面有m个P，后面有n个T，那么以这个A为中心形成的PAT个数为mn个。统计所有的即可，P[i]表示第i个字符前有几个P，T[i]表示第i个字符后有几个T。
注意要取模1000000007。

代码

#include "iostream"#include "string"using namespace std;string s;long long P[100000 + 5];long long T[100000 + 5];long long ans = 0;void init(){    cin>>s;    int i;    P[0] = 0;    for(i = 1; i < s.length(); i++){        P[i] = (s[i-1] == 'P')?P[i - 1] + 1:P[i - 1];    }    T[s.length() - 1] = 0;    for(i = s.length() - 2; i >= 0; i--){        T[i] = (s[i+1] == 'T')?T[i + 1] + 1:T[i + 1];    }}void cal(){    int i;    for(i = 1; i < s.length() - 1; i++){        if(s[i] == 'A')            ans += P[i] * T[i] % 1000000007;    }}int main(){    init();    cal();    cout<<ans % 1000000007<<endl;    system("pause");}