pat1015. Reversible Primes (20)

1015. Reversible Primes (20)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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

---

提交代码

 #include <cstdio>
 #include <cstring>
 #include <string>
 #include <queue>
 #include <stack>
 #include <iostream>
 using namespace std;
 bool prime[];
 void getprime(int n){
     memset(prime,false,sizeof(prime));
     int i,j;
     prime[]=true;
     for(i=;i<=n;i+=){
         prime[i]=true;
         for(j=;j*j<=i;j++){
             if(i%j==){
                 prime[i]=false;
                 break;
             }
         }
     }
 }
 int main(){
     //freopen("D:\\INPUT.txt","r",stdin);
     getprime();
     int n,d;
     while(scanf("%d",&n)!=EOF){
         if(n<){
             break;
         }
         scanf("%d",&d);
         if(!prime[n]){
             printf("No\n");
             continue;
         }
         queue<int> q;
         while(n){
             q.push(n%d);
             n/=d;
         }
         //cout<<n<<endl;
         while(!q.empty()){
             n*=d;
             n+=q.front();
             q.pop();
         }
         //cout<<n<<endl;
         if(prime[n]){
             printf("Yes\n");
         }
         else{
             printf("No\n");
         }
     }
     return ;
 }