1015 Reversible Primes (20)

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本文介绍了一种判断特定进制下可逆素数的方法。通过输入两个正整数N和D，程序会判断N在D进制下是否既是素数且其逆也是素数。文章提供了完整的C++代码实现，并详细解释了素数判断及进制转换的过程。

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1015 Reversible Primes (20)
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 (< 10^5^) 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

解题思路：
对素数的判断。
只要输入的n为正，就持续进行；先判断原n是否为素数，否就直接输出No。将n转化为d进制数，再将其反转，转化为十进制，判断是否为素数。

#include<cstdio>
#include<cmath> 
#include<algorithm>
using namespace std;
bool isprime(int n){
    if(n<=1) return false;
    int sqr=(int)sqrt(1.0*n);
    for(int i=2;i<=sqr;i++){
        if(n%i==0) return false;
    }
    return true;
}
int change(int n,int d){

    return n;
}
int main(){
    int n,d;
    while(scanf("%d",&n)!=EOF){
        if(n<0) break;
        scanf("%d",&d);
        if(isprime(n)==false) printf("No\n");
        else{
            int index=0,ans[100];
            do{
                ans[index++]=n%d;
                n/=d;
            }while(n);
            for(int i=0;i<index;i++) {
                n=n*d+ans[i];
            }
            if(isprime(n)==false) printf("No\n");
            else printf("Yes\n");
        }
    }
    return 0;
}