题目描述

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.

输入描述

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.

输出描述

For each test case, print in one line Yes if N is a reversible prime with radix D, or No if not.

输入例子

73 10
23 2
23 10
-2

输出例子

Yes
Yes
No

我的代码

#include<stdio.h>
#include<math.h>
#include<iostream>
using namespace std;
struct A{
    int num;
    int r;
}a[100010];

int getRadix(int n,int r){
    int ans=0;
    int k[10010]={-1},i=0;
    while(n!=0){
        k[i]=n%r;
        n=n/r;
        i++;
    }
    for(int j=0;j<i;j++){
        ans=ans+k[j]*pow(r,i-1-j);
    }
    return ans;
    
}

bool isPrime(int n){
    int sqr=(int)sqrt(1.0*n);
    for(int i=2;i<=sqr;i++){
        if(n%i==0){
            return false;
        }
    }
    return true;
}


int main(){
    int count=0;
    while(1){
        scanf("%d",&a[count].num);
        if(a[count].num<0){
            break;
        }
        scanf("%d",&a[count].r);
        count++;
    }
    int rev;
    for(int i=0;i<count;i++){
        rev=getRadix(a[i].num,a[i].r);
        if(a[i].num==1){
            printf("No\n");
            continue; 
        }
        if(isPrime(a[i].num) && isPrime(rev)){
            printf("Yes\n");
        }
        else{
            printf("No\n");
        }
    }
    
    return 0;
}