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 (<) and D (1), 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

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<cstring>
#include<string.h>
using namespace std;
const int maxn=110010;

int isR(int n,int d){
    int m=0;
    int a[maxn];
    int i=0;
    while(n>0){
        a[i]=n%d;
        n/=d;
        i++;
    }
    for(int j=0;j<=i-1;j++){
        m=m*d+a[j];
    }
    return m;
}

bool isPrime(int n){
    if(n==0){
        return false;
    }
    if(n==1){    //1既不是素数也不是合数 
        return false;
    }
    for(int i=2;i<=sqrt(n);i++){    //注意：i<=sqrt(n) 否则4判错 
        if(n%i==0){
            return false;
        }
    }
    return true;
}

int main(){
    int n,d;
    int a,b;
    while(scanf("%d",&n)){
        a=0;
        b=0;
        if(n<0){
            return 0;
        }
        scanf("%d",&d);
        if(isPrime(n)){
            a=1;
        }
        n=isR(n,d);
        if(isPrime(n)){
            b=1;
        }
        if(a==1&&b==1){
            printf("Yes\n");
        }else{
            printf("No\n");
        }
    }
    return 0;
}