PAT (Advanced Level) Practise 1015. Reversible Primes (20)

最新推荐文章于 2018-01-15 09:14:15 发布

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最新推荐文章于 2018-01-15 09:14:15 发布

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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，并提供了一个简洁的C++实现示例。通过输入任意正整数N和进制D，程序能够判断N在D进制下是否为可逆素数。

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 (< 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:

Sample Output:

Yes
Yes
No

题解：

这道题比较简单，先判断原来的数在D进制下是否为素数，如果不是素数，则直接输出No;如果是素数，则进一步判断该数在D进制下的逆序数是否为素数。

代码：

#include <cstdio>
#include <cmath>
int is_prime(int n){
    if(n<=1) return 0;
    int m=round(sqrt(n));
    for(int i=2;i<=m;i++){
        if(n%i==0) return 0;
    }
    return 1;
}

int main(){
    int n,d;
    while(scanf("%d%d",&n,&d)==2){
        if(n<0) break;
        if(!is_prime(n)) puts("No");
        else{
            int a[40],num=0,t=n;
            do{
                a[num++]=n%d;
                n/=d;
            }while(n);
            int sum=0;
            for(int i=0;i<num;i++) sum=sum*d+a[i];
            if(is_prime(t)&&is_prime(sum)) puts("Yes");
            else puts("No");
        }
    }
    return 0;
}