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 (<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:

73 10
23 2
23 10
-2

Sample Output:

Yes
Yes
No

一开始以为题目设置反向取值是为了设坑，但实际上是为了方便我们计算，在十进制转其他进制时就不用反取余数了

#include<bits/stdc++.h>
using namespace std;
int pow(int a,int n){  
    int x=1;
    for(int i=0;i<n;++i){
        x*=a;
    }
    return x;
}
bool JudgePrime(int x){  //判断质数
    if (x==1){
        return false;
    }
    else if (x==2){
        return true;
    }
    for (int i=2;i<=x/2+1;++i){
        if (x%i==0){
            return false;
        }
    }
    return true;
}
int DtoX(int d,int x){  //转x进制后反向取值
    int s[100];
    int out=0;
    int i=0;
    
    while(d){
        s[i]=d%x;
        d=d/x;
        ++i;
    }
    int q=0;
    for (int j=i-1;j>=0;--j,++q){
        out+=s[j]*pow(x,q);
    }
    return out;
}
int main(){
    int n,d;

    while(cin>>n&&n>0){
        cin>>d;
        if (JudgePrime(n)&&JudgePrime(DtoX(n,d)))
            cout<<"Yes"<<endl;
        else
            cout<<"No"<<endl;
    }
}