1015. Reversible Primes (20)

最新推荐文章于 2022-02-22 19:31:52 发布

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最新推荐文章于 2022-02-22 19:31:52 发布

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本文介绍了一种算法，用于判断一个十进制正整数N是否为D进制下的可逆素数。通过将N转换为D进制并反转其位数再转回十进制，检查原始数及反转后的数是否均为素数。文章提供了完整的C++实现代码。

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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

N是十进制的，化为D进制存入数组，反一下后再化为十进制判断是否为素数。

#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;
bool isp(int x)
{
    if (x<2) return false;
    for (int i=2;i<=sqrt(x);i++)
        if (x%i==0) return false;
    return true;

}
int main()
{
    int n,m,i,j,k;
    int a[33],b[33];
    while (true)
    {
        scanf("%d",&n);
        if (n<0) break;
        scanf("%d",&m);
        int cnt=0;
        k=n;
        while (k)
        {
            a[cnt++]=k%m;
            k/=m;
        }
        int t=0;
        for (i=cnt-1;i>=0;i--)
            b[t++]=a[i];
        int now=1;int sum=0;
        for (i=0;i<t;i++)
        {
            sum=b[i]*now+sum;
            now*=m;
        }
        if (isp(n)&&isp(sum))
            printf("Yes\n");
        else printf("No\n");
    } 
    return 0;
}