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"); }

转载于:https://www.cnblogs.com/z-y-k/p/11565662.html