PTA甲级——1093

1093 Count PAT’s

原题

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:

1
APPAPT

Sample Output:

1
2

思路

​ 大致题意：寻找子序列为PAT的数量，这里的子序列可以不是连续的，看案例可知。

​ 思路：就是用2个前缀和来分别存字符P在各个位置从前往后看的数量以及T在各个位置从后往前看的数量就ok了。

代码

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#include <iostream>
using namespace std;
const int N = 1e5 + 10, mod =  1000000007;
int n;
int P[N], T[N], TT[N];
long long res;

int main()
{
    string s;
    cin >> s;
    s = " " + s;
    int len = s.length();
    for (int i = 1; i < len; i++) {
        P[i] = P[i - 1];
        T[i] = T[i - 1];
        if (s[i] == 'P') {
            P[i]++;
        } else if (s[i] == 'T') {
            T[i]++;
        }
    }

    for (int i = 1; i < len; i++) {
        TT[i] = T[len - 1] - T[i - 1];
        // cout << T[i] << '\n';
    }
    for (int i = 1; i <= s.length(); i++) {
        if (s[i] == 'A') {
            res = (res + P[i] * TT[i]) % mod;
        }
    }

    cout << res;
}