pat 1015. 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

解：卡了，醉了，卡1：需要判断n值前后都是质数，第一次我只判断n值逆转后的了，很快就补上了；卡2：判断prime(n)==1&&prime(res)==1，我没注意n值已经换了，不是最初输出的值了，还需仔细。

代码：

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
using namespace std;
int prime(int n)
{
    if(n==1) return 0;
    if(n==2||n==3||n==5||n==7) return 1;
    int k=floor(sqrt(n)+0.5);
    for(int i=2;i<=k;i++)
    {
        if(n%i==0)
            return 0;
    }
    return 1;
}
int main()
{
    int n,m;
    while(scanf("%d",&n)==1&&n>0)
    {
        scanf("%d",&m);
        int wa=n;
        if(prime(n)==0)
        {
             printf("No
");
             continue;
        }
        int a[20],cnt=0;
        while(n)
        {
            a[cnt++]=n%m;
            n=n/m;
        }
        int res=0,c=0;
        for(int i=cnt-1;i>=0;i--)
        {
            int temp=1;
            for(int j=0;j<c;j++) temp*=m;
            c++;
            res+=a[i]*temp;
        }
        if(prime(wa)==0||prime(res)==0)
            printf("No
");
        else
            printf("Yes
");
    }
}

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