1015. Reversible Primes (20)

题目连接：https://www.patest.cn/contests/pat-a-practise/1015

题目如下：

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

这道题要得揣摩出输入的N是十进制的……
大致有两种方法，一种是将输入的数Ｎ不断除以基数，并将每次得到的余数存入数组中，然后计算得出反转后的数，判断即可；
还有一种是推到出迭代公式，这个公式我实在是自己推不出来，希望哪位网友能帮忙回答下这个迭代公式是怎么来的……
代码一（数组）：

#include<stdio.h>
#include<math.h>
#define MAXN 100005

int rev[MAXN];
int IsPrime(int num)
{
    if (num<2)return 0;
    else if (num==2 ||num==3)return 1;
    if (num%2==0)return 0;
    int i;
    for (i=3;i<=sqrt(num);i=i+2)
    {
        if (num%i==0) return  0;
    }
    return 1;
}

int main()
{
    int num,d,i=0,ori,result=0;
    scanf("%d",&num);
    while (num>=0)
    {
        ori=num;
        scanf("%d",&d);
        int radix=1;
        do
        {
            rev[i++]=num%d;
            num/=d;
        }while (num);
        for (--i;i>=0;i--)
        {
            result+=radix*rev[i];
            radix*=d;
        }
        if (IsPrime(ori) && IsPrime(result))printf("Yes\n");
        else printf("No\n");
        result=0;
        i=0;
        scanf("%d",&num);
    }
}

代码二（迭代公式）：

#include<stdio.h>
#include<math.h>

int IsPrime(int num)
{
    if (num<2)return 0;
    if (num==2 || num==3)return 1;
    if (num%2==0)return 0;
    int i;
    for (i=3;i<=sqrt(num);i+=2)
    {
        if (num%i==0)return 0;
    }
    return 1;
}

int main()
{
    int num,d,ori,result=0;
    scanf("%d",&num);
    while (num>=0)
    {
        scanf("%d",&d);
        ori=num;
        if (IsPrime(ori))
        {
            do
            {
                result=result*d+num%d;
                num/=d;
            }while(num);
            if (IsPrime(result))printf("Yes\n");
            else printf("No\n");
        }
        else printf("No\n");
        scanf("%d",&num);
        result=0;
    }
    return 0;
}