1015. Reversible Primes (20)

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

题目大意：判断原数字是否是素数，倒序后是否是素数，如果都是素数输出Yse

                 这里的是指原数字与它在数制下的倒序，如23 2，是10111（2）的倒叙11101的真实数字。

#include<stdio.h>
#include<math.h>
#include<stdlib.h>
int in[6];
int main()
{
    int n,d,m,temp;
    int i,j;
    int sum,flag;
    while( scanf("%d",&n)!=EOF)
    {
        if( n<0 ) break;
        scanf("%d",&d);
        i=0;  //分解整数的下标
        sum=0;  //转换后的值
        flag = 0;  //标记是否能转换
        temp=n;   //保存n的值
        
        while( n ) //分解位数
        {
            in[i++]=n%d;
            n /= d;
        }
        for( j=0; j<i ; j++) //求转换的值
            sum = sum*d+ in[j];
        if( isPrime(sum) && isPrime(temp)) //如果转换前和转换后都是素数
            flag =1;
        if( !flag) printf("No
");
        else printf("Yes
");
    }
    return 0;
}

int isPrime( int sum)
{
    int i,m;
    if( sum<2 ) return 0;
    if( sum==2 || sum ==3 )return 1;
    m =(int) sqrt(sum)+1;
    for( i=2; i<m; i++)
        if( sum%i==0 )
            return 0;
    return 1;
}