PAT1093: Count PAT's

1093. Count PAT's (25)

时间限制

120 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CAO, Peng

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

试了下直接暴搜，果然最后几个测试点超时，看来还是得用dp。
这道题DP的思路就是用两个int数组P[n],T[n]保存第i个字符的左边字母'P'的个数P[i]和右边字母'T'的个数T[i]，当遍历到的字符是'A'时，那么以这个A为基础的"PAT"字符串个数就是它左侧P字母个数乘以右侧T字母个数。即P[i]*T[i]

代码

#include<iostream>
#include<vector>
using namespace std;
int main()
{
    string s;
    while(cin >> s)
    {
        vector<int> P(100000,0);
        vector<int> T(100000,0);
        for(int i = 1,j = s.size() - 2;i < s.size() && j >= 0;i++,j--)
        {
            if(s[i - 1] == 'P')
            {
                P[i]++;
            }
            if(s[j + 1] == 'T')
            {
                T[j]++;
            }
            P[i] += P[i - 1];
            T[j] += T[j + 1];
        }

        int sum = 0;
        for(int i = 0;i < s.size();i++)
        {
            if(s[i] == 'A')
            {
                sum += P[i] * T[i];
                sum %= 1000000007;
            }

        }
        cout << sum << endl;
    }
}