PAT甲级A1015 Reversible Primes (20)

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:

73 10
23 2
23 10
-2

Sample Output:

Yes
Yes
No

题意：给定一个正整数n，和一个进制d，如果n在10进制下是素数,并且将其d进制逆转之后再转换成10进制数,仍然是素数，则输出Yes，否则输出No。

思路：自己编写两个进制转换函数，以及素数判断函数按要求操作即可。

参考代码：

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#include<cstdio>
#include<iostream>
#include<string>
#include<cmath>
using namespace std;
bool judge(int n)
{
	if(n<=1) return false;
	int sqrtN=(int)sqrt(1.0*n);
	for(int i=2;i<=sqrtN;i++)
	{
		if(n%i==0)
			return false;
	}
	return true;
}
string intToA(int n,int radix)
{
	string ans="";
	do{
		ans+=n%radix+'0';	
		n/=radix;
	}while(n!=0);	//使用do{}while（）以防止输入为0的情况
	return ans;	//此时的字符串正好存储逆置后的结果
}
int main()
{
	int n,radix;
	while(cin>>n,n>=0)
	{
		cin>>radix;
		string s;
		if(!judge(n)) {
			printf("No\n");
			continue;
		}
		s=intToA(n,radix);	
		n=0;
		for(int i=0;i<s.size();i++)
		{
			n=n*radix+s[i]-'0';
		}
		if(judge(n)) printf("Yes\n");
		else printf("No\n");
	}
	return 0;
}