PAT A 1093. Count PAT's (25)

题目

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

典型的dp问题，

到s[j]时，匹配到“PAT”的第i个元素位置时的匹配数表示为：num[i][j]，则：

1、如果s[j]等于PAT[i]，num[i][j]=num[i][j-1]+num[i-1][j]；

2、否则，num[i][j]=num[i][j-1]。

对于取余为防止溢出，每当为情况1时取一次余即可。

代码：

#include <iostream>
#include <algorithm>
#include <string>
#include <vector>
using namespace std;

int main()
{
	string s;
	cin>>s;

	if(s.empty())
	{
		cout<<0;
		return 0;
	}
	char cf[3]={'P','A','T'};
	vector<vector<int>> num(3,vector<int>(s.size(),0));	//num[i][j]表示到s[j]时匹配目标串cf[0~i]的匹配数
	if(s[0]=='P')	//dp
		num[0][0]=1;
	for(int i=1;i<s.size();i++)
		if(s[i]=='P')
			num[0][i]=num[0][i-1]+1;
		else
			num[0][i]=num[0][i-1];
	for(int i=1;i<3;i++)
		for(int j=1;j<s.size();j++)
			if(s[j]==cf[i])
			{
				num[i][j]=num[i][j-1]+num[i-1][j];
				num[i][j]%=1000000007;
			}
			else
				num[i][j]=num[i][j-1];

	cout<<num[2][s.size()-1];

	return 0;
}