【PTA-A】1015 Reversible Primes（判断素数、进制转换、进制逆序）

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本文介绍了一个算法，用于判断一个正整数在特定进制下是否为可逆素数，即该数及其在指定进制下的逆序数均为素数。通过进制转换和素数判断，实现了对输入数字的有效验证。

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 (<105) 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

思路：

1.先判断原数字是不是素数

2.进制反转后的数是不是素数

注意点：

1.1不是素数

2.输入小于0时结束

#include<iostream>
#include<algorithm>
#include<cmath>
using namespace std;
int d[100001];
bool isprime(int n) {
	if (n <= 1)return false;
	int t = int(sqrt(n*1.0));
	for (int i = 2; i <= t; i++) {
		if (n % i == 0)return false;
	}
	return true;
}
int main() {
	int n, radix;
	while (scanf("%d",&n) != EOF&& n >= 0) {
		cin >> radix;
		if (!isprime(n)) {
			cout << "No" << endl;
		}else{
			//进制转换
			int len = 0;
			while (n != 0){
				d[len++] = n % radix;
				n /= radix;
			} 
			//进制逆序
			for (int i = 0; i < len; i++) {
				n = n * radix + d[i];
			}
			if (isprime(n))cout << "Yes"<<endl;
			else cout << "No"<<endl;
		}
	}
	return 0;
}