PAT1015

A reversible prime in any number system is a prime whose "reverse" in that number system is also a prime.

一个反素数是任意数字系统中，不仅本身是素数反的也是素数

For example in the decimal system 73 is a reversible prime because its reverse 37 is also a prime.

比如在小数系统中73是一个反素数，因为37也是素数。

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.

现在给出任意两个正整数N和D，要求你写出N的D进制是不是反素数

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

判断N是不是素数，然后将N转换成D进制的数，然后翻转，然后转成10进制，然后判断结果和N是不是都是素数。

如23 2

转换成2进制是10111

反转是11101

转换成十进制是29

因为29和23都是素数，所以输出yes

#include<cstdio>
#include<cstdlib>
#include<iostream>
#include<algorithm>

using namespace std;

//判读是不是素数，是返回1，不是返回0
int isSu(int number)
{
    if(number < )
        return ;
    for (int i = ; i*i <= number; i++)
    {
        if(number%i == )
            return ;
    }
    return ;
}

//将一个整数转换成任意进制,，逆序之后转换成10进制返回
int change(int a,int d)
{
    int result=;
    int q=;
    int jinzhi[];
    while (a>)
    {
        jinzhi[q] = a%d;
        a /= d;
        q++;
    }
    int dishu = ;
    for (int i = q-; i >= ; i--)
    {
        result += jinzhi[i]*dishu;
        dishu*=d;
    }
    return result;
}

int main()
{
    int n,d;
    while (true)
    {
        cin>>n;
        if(n<)
            break;

        cin>>d;

        if(isSu(n) && isSu(change(n,d)))
            cout<<"Yes"<<endl;
        else
            cout<<"No"<<endl;
    }
    return ;
}

两点值得说明的，素数的判断没有用sqrt（）而是用了i*i,还有就是我第一次错在，1不是素数上面了，其他都特别简单