1015. Reversible Primes (20)

最新推荐文章于 2024-02-22 13:10:55 发布

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最新推荐文章于 2024-02-22 13:10:55 发布

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本文介绍了一种判断可逆素数的方法，即在一个特定的基数系统中，如果一个数及其反转都是素数，则该数为可逆素数。文章通过示例解释了可逆素数的概念，并提供了一个C++实现，包括素数判断和进制转换的功能。

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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:

Sample Output:

Yes
Yes
No

题解：

素数判别法。进制转换。

代码：

#include <cstdio>
#include <cmath>

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

int reverse(int n, int d)
{
	int ret = 0;
	while(n)
	{
		ret = ret * d + n % d;
		n /= d;
	}
	return ret;
}
int main()
{
	int n, radix;
	while(scanf("%d", &n) != EOF)
	{
		if(n < 0)
			break;
		scanf("%d", &radix);
		int rev = reverse(n, radix);
		if(isPrime(rev) && isPrime(n))
			printf("Yes\n");
		else
			printf("No\n");
	}
	
	return 0;

}