PAT(进制转换)——1015 Reversible Primes （20 分）

1015 Reversible Primes （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:

73 10
23 2
23 10
-2

Sample Output:

Yes
Yes
No

题目大意：

如果一个数本身是素数，并且转换为以d为基数反转过来仍然是素数，则输出yes。

题目解析：

一开始理解错了题意，翻转过来还是要以d为基数翻转。

具体代码：

#include<iostream>
#include<algorithm>
#include<cstring>
#include<cmath>
using namespace std;
#define N 11000
bool isprime(int n) {
    if(n <= 1) return false;
    int sqr = int(sqrt(n * 1.0));
    for(int i = 2; i <= sqr; i++) {
        if(n % i == 0)
            return false;
    }
    return true;
}
int convert(string n,int d){
	int num=0;
	for(int i=0;i<n.size();i++){
		num=d*num+(n[i]-'0');
	}
	return num;
}
int resort(int n){
	int num=0;
	while(n!=0){
		num=num*10+n%10;
		n/=10;
	}
	return num;
}

int main()
{
    int n;
	int d;
    while(cin>>n){
    	if(n<0) break;
    	cin>>d;
    	if(isprime(n)){
    		char A[20];
    		int top=0;
    		do{
    			A[top++]=n%d+'0';
			}while(n/=d);
			for(int i=0;i<top;i++){
				n=n*d+(A[i]-'0');
			}
    		if(isprime(n))
    			cout<<"Yes"<<endl;
    		else
    			cout<<"No"<<endl;
		}
    	else
    		cout<<"No"<<endl;
	}
    return 0;
}