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[100005];
void getprime(int n){
    memset(prime,false,sizeof(prime));
    int i,j;
    prime[2]=true;
    for(i=3;i<=n;i+=2){
        prime[i]=true;
        for(j=3;j*j<=i;j++){
            if(i%j==0){
                prime[i]=false;
                break;
            }
        }
    }
}
int main(){
    //freopen("D:\\INPUT.txt","r",stdin);
    getprime(100005);
    int n,d;
    while(scanf("%d",&n)!=EOF){
        if(n<0){
            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 0;
}