先贴题目：

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 1023 223 10-2

Sample Output:

YesYesNo 题解：本题意思是判断一个数和该数在规定进制条件下反转后是否为素数。题目本身比较简单，但是一定要理解题意（我一开始就理解错题意） 关键是求该数在规定进制反转后在转为十进制后的数是否为素数。举个例子： 23 2 这组答案，23在10进制下是个素数，转化为二进制为10111.反转后为11101，再转为10进制为39，也是个质数。所以输出Yes 题目思路： 首先需要判断某个数是否为素数的函数（注意二为素数） 其次将这个数按照位数反转即可。 代码如下：

1 #include 
     
       2 #include
      
        3 using namespace std; 4 int convert(int n,int r); 5 int convert_todicimal(int n,int r); 6 bool judge(int n) 7 { 8     int i; 9     if(n<=1)10         return false;11     if(n==2||n==3)12         return true;13     for(i=2;i<=sqrt(n)+1;i++)14     {15         if(n%i==0)16         {17             return false;18         }19     }20     return true;21 }22 int reverse(int n,int r)//十进制数倒置23 {24     int result=0;25     int temp=0;26     while(n!=0)27     {28         int t=n%r;29         result=(result*r)+t;30         n=n/r;31     }32     return result;33 }34 35 int main() {36     int n,r;37     while(1)38     {39         scanf("%d%d",&n,&r);40         if(n<0)41             break;42         //printf("%d\n",convert(n,r));43         //printf("%d\n",reverse(n,r));44         if(judge(n)&&judge(reverse(n,r)))45         {46             printf("Yes\n");47         }48         else49         {50             printf("No\n");51         }52     }53     return 0;54 }