【c++】PAT (Advanced Level)1015. Reversible Primes (20)

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本文介绍了一个用于检测任意基数下可逆质数的算法，详细解释了可逆质数的概念，并通过实例展示了如何实现该检测过程。

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1015. Reversible Primes (20)

时间限制

400 ms

内存限制

32000 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

A reversible prime in any number system is a prime whose "reverse" in that number system is also a prime. For example in the decimal system 73 is a reversible prime because its reverse 37 is also a prime.

Now given any two positive integers N (< 10⁵) and D (1 < D <= 10), you are supposed to tell if N is a reversible prime with radix D.

Input Specification:

The input file consists of several test cases. Each case occupies a line which contains two integers N and D. The input is finished by a negative N.

Output Specification:

For each test case, print in one line "Yes" if N is a reversible prime with radix D, or "No" if not.

Sample Input:

Sample Output:

Yes
Yes
No

啊啊啊，浙大的PAT 停了

1.格式错误，因为忘了加回车了

2.一个测试点是关于2 的，一个测试点是关于1 的

#include <iostream>
#include <string>
#include <vector>
#include <iomanip>
#include <algorithm>
using namespace std;
//C7
bool isPrime(int m){
	int j;  
	bool flag=true;  
	for(j=2;j*j<=m;j++){  
		if(m==2)  
			break;  
		else if(m%j==0){  
			flag=false;  
		}  

	}  

	if(m==1){
		flag=false;
	}
	return flag;  
}



int main() {
	while(true){
		int num,num1;
		cin>>num;
		num1=num;
		if(num<0){
			break;
		}
		while(num>0){
			int radix,fix=1,sum=0;
			cin>>radix;
			while(num>0){
				fix=num%radix;
				sum=sum*radix+fix;
				num=num/radix;

			}
			//	cout<<sum;
			bool flag;
			if(isPrime(num1)&&isPrime(sum)){
				cout<<"Yes"<<endl;
			}else{
				cout<<"No"<<endl;
			}
		}
	}
	return 0;
}