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

一開始以為題目設定反向取值是為了設坑，但實際上是為了方便我們計算，在十進位制轉其他進位制時就不用反取餘數了

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 int pow(int a,int n){  
 4     int x=1;
 5     for(int i=0;i<n;++i){
 6         x*=a;
 7     }
 8     return x;
 9 }
10 bool JudgePrime(int
 
 x){  //判斷質數
11     if (x==1){
12         return false;
13     }
14     else if (x==2){
15         return true;
16     }
17     for (int i=2;i<=x/2+1;++i){
18         if (x%i==0){
19             return false;
20         }
21     }
22     return true;
23 }
24 int DtoX(int d,int x){  //轉x進位制後反向取值
25     int s[100];
26     int out=0;
27     int i=0;
28     
29     while(d){
30         s[i]=d%x;
31         d=d/x;
32         ++i;
33     }
34     int q=0;
35     for (int j=i-1;j>=0;--j,++q){
36         out+=s[j]*pow(x,q);
37     }
38     return out;
39 }
40 int main(){
41     int n,d;
42 
43     while(cin>>n&&n>0){
44         cin>>d;
45         if (JudgePrime(n)&&JudgePrime(DtoX(n,d)))
46             cout<<"Yes"<<endl;
47         else
48             cout<<"No"<<endl;
49     }
50 }