PAT-1015 Reversible Primes (20 分) 进制转换+质数

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

注意读题,首先他得是一个质数才行,这个先决条件我一开始没有判断,另外判断质数时要注意1,2,3和负数的情况

#include<bits/stdc++.h>
#define de(x) cout<<#x<<" "<<(x)<<endl
#define each(a,b,c) for(int a=b;a<=c;a++)
using namespace std;
const int maxn=50+5;
int buf[maxn];
bool isprime(int x)
{
    if(x<=1)return false;
    bool flag=true;
    if(x==2)return true;
    if(x==3)return true;
    for(int i=2;i*i<=x;i++)
    {
        if(x%i==0)
        {
            flag=false;
            break;
        }
    }
    return flag;
}
int main()
{
    //de(isprime(73));
    int n,r;
    while(true)
    {
        cin>>n;
        if(n<0)break;
        cin>>r;
        if(n==1||n==0)
        {
            puts("No");
            continue;
        }
        if(!isprime(n))///读题的问题,首先得判断他是不是一个质数才行,
        {
            puts("No");
            continue;
        }
        int len=0;
        while(n>0)
        {
            buf[len++]=n%r;
            n/=r;
        }
        //de(len);
        int sum=0;
        len--;
        for(int i=0;i<=len;i++)
        {
            sum+=buf[i]*pow(r,len-i);
        }
        //de(sum);
        if(isprime(sum))puts("Yes");
        else puts("No");
    }
}