1015 Reversible Primes (20)

最新推荐文章于 2020-06-23 18:08:06 发布

转载最新推荐文章于 2020-06-23 18:08:06 发布 · 145 阅读

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原文链接：http://www.cnblogs.com/Yaxadu/p/9164459.html

本文介绍了一种算法，用于判断一个给定的正整数是否为特定进制下的可逆素数。可逆素数是指在一个数系中，该数及其在该数系中的反转形式均为素数的情况。文章提供了完整的C语言实现代码，并通过样例输入输出展示了程序的功能。

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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:

Sample Output:

Yes
Yes
No

#include<stdio.h>
#include<cmath>

bool isPrime(int n) {
    if (n <= 1) return false;
    int sqr = (int) sqrt(1.0 * n);
    for (int i = 2; i <= sqr; ++i) {
        if (n % i == 0) return false;
    }
    return true;
}
int d[111];
int main() {
    int n, radix;
    while (scanf("%d", &n) != EOF) {
        if (n < 0) break;
        scanf("%d", &radix);
        if (isPrime(n) == false) printf("No\n");
        else {
            int len = 0;
            do {
                d[len++] = n % radix;
                n /= radix;
            }while (n != 0);
            for (int i = 0; i < len; i++) {
                n = n * radix + d[i];
            }
            if (isPrime(n) == true) printf("Yes\n");
            else printf("No\n");
        }
    }
    return 0;
}