浙大pat甲级题目---Reversible Primes (20)

先贴题目：

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

题解：本题意思是判断一个数和该数在规定进制条件下反转后是否为素数。题目本身比较简单，但是一定要理解题意（我一开始就理解错题意）
关键是求该数在规定进制反转后在转为十进制后的数是否为素数。举个例子：
23 2 这组答案，23在10进制下是个素数，转化为二进制为10111.反转后为11101，再转为10进制为39，也是个质数。所以输出Yes
题目思路：
首先需要判断某个数是否为素数的函数（注意二为素数）
其次将这个数按照位数反转即可。
代码如下：

#include <iostream>
#include<math.h>
using namespace std;
int convert(int n,int r);
int convert_todicimal(int n,int r);
bool judge(int n)
{
    int i;
    if(n<=1)
        return false;
    if(n==2||n==3)
        return true;
    for(i=2;i<=sqrt(n)+1;i++)
    {
        if(n%i==0)
        {
            return false;
        }
    }
    return true;
}
int reverse(int n,int r)//十进制数倒置
{
    int result=0;
    int temp=0;
    while(n!=0)
    {
        int t=n%r;
        result=(result*r)+t;
        n=n/r;
    }
    return result;
}

int main() {
    int n,r;
    while(1)
    {
        scanf("%d%d",&n,&r);
        if(n<0)
            break;
        //printf("%d\n",convert(n,r));
        //printf("%d\n",reverse(n,r));
        if(judge(n)&&judge(reverse(n,r)))
        {
            printf("Yes\n");
        }
        else
        {
            printf("No\n");
        }
    }
    return 0;
}