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) // 遮蔽VS2017對scanf和prinf的報錯

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進位制轉d進位制
		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

本程式碼是來自柳婼大佬的，本來自己打了一遍，結果卡在第二個測試點，看來平時覺得很簡單的程式碼還有些細節沒有注意，現在學習了，以後對於判斷素數就這麼寫了

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;
}

對於10進位制轉d進位制就這麼寫（這個是倒轉的）

// 10進位制轉d進位制
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