PAT 甲级 1015. Reversible Primes

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本文介绍了一个编程问题，即如何判断一个十进制或特定进制下的数字是否为可逆素数，即该数字及其翻转后形成的数字均为素数。文章提供了完整的C++代码实现，包括素数判断及进制转换的方法。

1015. Reversible Primes (20)

时间限制

400 ms

内存限制

65536 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

题意：给你两个数，n表示需要的数字，d表示进制。要让n变为d进制后的数字反转一遍再变成十进制，如果他们两个数都是素数，就YES，不然就NO

#include<iostream>
using namespace std;
bool judge(int n){
	if(n<2) return false;
	if(n==2||n==3) return true;
	for(int i=2;i*i<=n;i++){
		if(n%i==0) return false;
	}
	return true;
}
int change(int num,int radix){
	int tmp=0;
	while(num){
		tmp*=radix;
		tmp+=(num%radix);
		num/=radix;
	}
	return tmp;
}
int main(){
	int n,d;
	while(cin>>n){
		if(n<0) break;
		cin>>d;
		int tmp=change(n,d);
		if(judge(n)&&judge(tmp)){
			cout<<"Yes"<<endl;
			continue;
		}
		else cout<<"No"<<endl;
	}
}