PTA A1015

A1015 Reversible Primes (20 分)

题目内容

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≤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

单词

reversible

英 /rɪ'vɜːsɪb(ə)l/ 美 /rɪ'vɝsəbl/
n. 双面布料
adj. 可逆的；可撤消的；可反转的

Prime

n. （美、法）普赖姆（人名）
adj. (prime) 主要的，首要的；最好的；典型的；最适宜的；素数的
n. (prime) 盛年，鼎盛时期
v. (prime) 使准备好

题目分析

弄清楚题目意思（这是我目前最大的困难之一，英文真的很难搞啊），这道题目还是很简单的，一是如何判断一个数是不是质数，而是如何计算出根据机制翻转后的数，由于之前做过一道和进制有关的题，所以做起来感觉没什么困难，对了，1不是质数，>_<。

具体代码

#include<stdio.h>
#include<stdlib.h>

int is_prime(int n)
{
    if(n==1)
    return 0;
	for (int i = 2; i < n; i++)
	{
		if (n%i == 0)
			return 0;
	}
	return 1;
}

int main(void)
{
	while (1)
	{
		int n, radix, rn = 0;
		scanf("%d", &n);
		if (n < 0)break;
		else
		{
			scanf("%d", &radix);
			if (is_prime(n))
			{
				while (n != 0)
				{
					int m = n % radix;
					rn = rn * radix + m;
					n /= radix;
				}
				if (is_prime(rn))
				{
					printf("Yes\n");
				}
				else
					printf("No\n");
			}
			else printf("No\n");
		}
	}
	system("pause");
}