【PAT】【Advanced Level】1093. Count PAT's (25)

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本文介绍了一种算法，用于计算给定字符串中特定子串“PAT”的出现次数。通过预先统计字符串中的字符分布，该算法能在O(n)的时间复杂度内解决问题，并提供了完整的C++实现代码。

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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:

原题链接：

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

思路：

对于每个A，能组成（前边P的个数）*（后边T的个数）个 PAT

先统计P,T，再循环一遍A即可

CODE：

#include<iostream>
#include<cstring>
#include<string>
#define N 100010
using namespace std;
int p[N];
int t[N];
int main()
{
	string str;
	p[0]=0;
	t[0]=0;
	cin>>str;
	for (int i=0;i<str.length();i++)
	{
		if (str[i]=='P')
		{
			p[i]=p[i-1]+1;
			t[i]=t[i-1];
		}
		else if (str[i]=='T')
		{
			p[i]=p[i-1];
			t[i]=t[i-1]+1;
		}
		else
		{
			p[i]=p[i-1];
			t[i]=t[i-1];
		}
		//cout<<p[i]<<" "<<t[i]<<endl;
	}
	int np=p[str.length()-1];
	int nt=t[str.length()-1];
	//cout<<np<<" "<<nt<<endl;
	long long sum=0;
	long long mo=1000000007;
	for (int i=0;i<str.length();i++)
	{
		if (str[i]=='A')
		{
			long long temp;
			//cout<<p[i]<<" "<<t[i]<<endl;
			temp=p[i]*(nt-t[i]);
			temp=temp%mo;
			//cout<<temp<<endl;
			sum=sum+temp;
			sum=sum%mo;
		}
	}
	cout<<sum;
	return 0;
}