A1015. Reversible Primes

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

 #include<cstdio>
 #include<iostream>
 #include<algorithm>
 #include<math.h>
 using namespace std;
 int numReverse(int N, int radix){
     int num[], temp, index = , ans = ;
     do{
         temp = N % radix;
         num[index++] = temp;
         N = N / radix;
     }while(N != );
     for(int i = index - , P = ; i >= ; i--){
         ans += num[i] * P;
         P = P * radix;
     }
     return ans;
 }
 int isPrime(int N){
     int sqr = (int)sqrt(N * 1.0);
     if(N == )
         return ;
     for(int i = ; i <= sqr; i++){
         if(N % i == )
             return ;
     }
     return ;
 }
 int main(){
     int N, D, N2;
     while(){
         scanf("%d", &N);
         if(N < )
             break;
         scanf("%d", &D);
         int temp = numReverse(N, D);
         if(isPrime(N) && isPrime(temp))
             printf("Yes\n");
         else printf("No\n");
     }
     cin >> N;
     return ;
 }

总结：

1、判断素数：

int isPrime(int N){
    int sqr = (int)sqrt(N * 1.0);  //N应该转换为小数
    if(N == )        //当N = 1时应返回false，容易忽略，1不是素数
        return ;
    for(int i = ; i <= sqr; i++){   //i <= sqr； i从2开始查找
        if(N % i == )
            return ;
    }
    return ;
}