PTA(Advanced Level)1015.Reversible Primes

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:

73 10
23 2
23 10
-2

Sample Output:

Yes
Yes
No

思路

题目的大意是判断一个数本身是否为素数，以及它在进制(D)下是否为素数

• 检查素数的方法和根据进制反转没有什么好讲的，挺简单的
• 值得一提的是本身不是素数就没必要再进行反转了

代码

#include<bits/stdc++.h>
using namespace std;
bool is_prime(int x)
{
	if(x <= 1)	return false;
	int up = (int)sqrt(x * 1.0);
	for(int i=2;i<=up;i++)
		if(x % i == 0)
			return false;
	return true;
}	//判断素数的方法

int getR(int n, int d)
{
	vector<int> v;
	int t = 0;
	while(n)
	{
		t = n % d;
		v.push_back(t);
		n /= d;
	}
	int r_value = 0;
	for(int i=0;i<v.size();i++)
	{
		r_value += v[i] * pow(d, v.size() - i - 1);
	}
	return r_value;
}   //根据进制得到反转之后的值

int main()
{
	int n, d;
	while(cin >> n)
	{
		if(n < 0)	break;
		else	cin >> d;
		if(!is_prime(n))	//如果本身不是素数，没有进一步检查的必要
		{
			cout << "No" << endl;
			continue;
		}
		else
		{
			int reverse_value = getR(n,d);
			if(is_prime(reverse_value))
				cout << "Yes" << endl;
			else cout << "No" << endl;
		}
	}
	return 0;
}

引用

https://pintia.cn/problem-sets/994805342720868352/problems/994805495863296000