PAT_A1015#Reversible Primes

Source:

PAT A1015 Reversible Primes (20 分)

Description:

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 (<) and D (1), 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 Nand 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

Keys:

• 进制转换
• 素数（Prime）

Attention:

• int < 1e9 ~ 2^8-1, long < 1e18 ~ 2^16-1；

• 1e5的十进制转换为二进制会超过int范围，进制转换和逆置一起进行的话就避免中间数超出范围的问题了；

Code:

/*
Data: 2019-05-12 21:32:05
Problem: PAT_A1015#Reversible Primes
AC: 18:26

题目大意：
给一整数N<1e5，进制D
求N的D进制逆置后是否仍是素数
*/
#include<cstdio>
#include<string>
#include<algorithm>
using namespace std;
const int M=1e5+10;

bool isPrime(int num)
{
    if(num==0 || num==1)
        return false;
    for(int i=2; i*i<=num; i++)
        if(num%i==0)
            return false;
    return true;
}

int Reverse(int n, int d)
{
    int ans=0;
    while(n!=0)
    {
        ans *= d;
        ans += (n%d);
        n /= d;
    }
    return ans;
}

int main()
{
#ifdef  ONLINE_JUDGE
#else
    freopen("Test.txt", "r", stdin);
#endif // ONLINE_JUDGE

    int n,d;
    while(scanf("%d",&n) && n>=0)
    {
        scanf("%d", &d);
        if(isPrime(n) && isPrime(Reverse(n,d)))
            printf("Yes\n");
        else
            printf("No\n");
    }

    return 0;
}