PAT甲级-1015 Reversible Primes (20)（20 分）【判断素数+反过来转化成其他进制】

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博客围绕可逆素数问题展开，给定两个正整数N和D，需判断N是否为基数D下的可逆素数。介绍了输入输出规格，还给出解题思路，如素数判断时i应小于等于sqrt(n)，反转用do - while更好，避免单独讨论等于0的情况。

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1015 Reversible Primes (20)（20 分）

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 (< 10^5^) 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

解题思路：

素数判断的时候，i<=sqrt（n）这句，我一开始交的时候没有打等于号，错了一个测试点，如果不能理解，那就把73转化成37时写的代码中把10改成d就好了。在反转用do-while比较好，这样等于0的时候就不用单独讨论了。

#include<bits/stdc++.h>
using namespace std;
int Reversible(int n,int d)
{
	int sum=0;
	do{
		sum=sum*d+n%d;
		n=n/d;
	}while(n!=0);
	return sum; 
}
int is_prime(int n)
{
	if(n==0||n==1)
	return 0;
	for(int i=2;i<=sqrt(n);i++)
	{
		if(n%i==0) return 0;
	}
	return 1;
}
int main(void)
{
	int n,d;
	while(~scanf("%d%d",&n,&d)&&n>=0)
	{
		//cout<<Reversible(n,d)<<endl;
		if(is_prime(n)&&is_prime(Reversible(n,d)))
		cout<<"Yes"<<endl;
		else
		cout<<"No"<<endl;
	}
	return 0;
}