PAT (Advanced Level) 1015 Reversible Primes （20 分）

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1015 Reversible Primes （20 分）

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

Code

#include <iostream>
#include <cmath>
#pragma warning(disable:4996) // 屏蔽VS2017对scanf和prinf的报错

using namespace std;

bool isprime(int n) {
	if (n <= 1) return false;
	int sqr = int(sqrt(n * 1.0));
	for (int i = 2; i <= sqr; i++) {
		if (n % i == 0)
			return false;
	}
	return true;
}
int main() {
	int n, d;
	while (scanf("%d", &n) != EOF) {
		if (n < 0) break;
		scanf("%d", &d);
		if (isprime(n) == false) {
			printf("No\n");
			continue;
		}
		int len = 0, arr[100];
		// 10进制转d进制
		do {
			arr[len++] = n % d;
			n = n / d;
		} while (n != 0);
		// finish
		for (int i = 0; i < len; i++)
			n = n * d + arr[i];
		printf("%s", isprime(n) ? "Yes\n" : "No\n");
	}
	return 0;
}

思路

本代码是来自柳婼大佬的，本来自己打了一遍，结果卡在第二个测试点，看来平时觉得很简单的代码还有些细节没有注意，现在学习了，以后对于判断素数就这么写了

bool isprime(int n) {
	if (n <= 1) return false;
	int sqr = int(sqrt(n * 1.0));
	for (int i = 2; i <= sqr; i++) {
		if (n % i == 0)
			return false;
	}
	return true;
}

对于10进制转d进制就这么写（这个是倒转的）

// 10进制转d进制
do {
	arr[len++] = n % d;
	n = n / d;
} while (n != 0);
// finish
for (int i = 0; i < len; i++)
	n = n * d + arr[i];

以上