1015 Reversible Primes

最新推荐文章于 2022-09-09 23:13:55 发布

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本文探讨了可逆素数的概念，即在任何数制中，一个素数的‘反转’也是素数。通过具体例子说明了如何判断一个给定的正整数是否为特定进制下的可逆素数，并提供了详细的算法实现，包括质数判断和进制转换的方法。

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 $N (<10^5)$ and $D (1 < D \leq 10)$ , 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:

Sample Output:

Yes
Yes
No

题解：

主要考察进制转换和质数判断，其中将十进制数转换为其他进制数时，按照竖式求解的方式本应是从下往上（即从左往右）取值的，但由于题中需要求其reverse，所以直接取值即可

#include <iostream>
#include <cmath>
using namespace std;

bool prime(int n){      // 质数判断
    if(n<2)   return false;
    if(n == 2 || n == 3) return true;
    int r = sqrt(n);
    for(int i=2;i<=r;i++){
        if(n%i == 0)
            return false;
    }
    return true;
}

int decimal_reverse(int num, int radix)		// 求其逆序的十进制表示
{
    int temp, sum = 0;
    while(num != 0){
        temp = num%radix;
        num /= radix;
        sum = sum*radix + temp;
    }
    return sum;
}

int main()
{
    while(1){
        int num, radix;
        cin>>num;
        if(num<0)
            break;
        cin>>radix;

        if(prime(num) && prime(decimal_reverse(num,radix)))
            cout<<"Yes\n";
        else
            cout<<"No\n";
    }
}