PAT甲级1015 Reversible Primes（简单模拟）

最新推荐文章于 2022-07-27 16:13:26 发布

threeh20

最新推荐文章于 2022-07-27 16:13:26 发布

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本文介绍了一种算法，用于判断一个数在特定进制下是否为可逆素数，即该数及其在该进制下的反转数均为素数。算法通过进制转换和素数验证实现，适用于不超过10^5的正整数和介于1到10之间的进制。

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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 (<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

求一个数在某个进制下（这里设置一个进制转换还硬核了。。。）是不是质数并且倒回来数是不是还是素数，按规则模拟一下就好。注意下细节问题。

#include<bits/stdc++.h>
using namespace std;
int main(){
	long long d,n;
	long long  a[111111];
	while(cin>>d)
	{
		if(d<0)break;
		cin>>n;
		int f1=0,f2=0;
		if(d==1)
		{
			cout<<"No"<<endl;
			continue;
		}
		for(long long i=2;i<=sqrt(d);i++)
		{
			if(d%i==0)
			f1=1;
		}
		int z=0;
		while(d)
		{
			a[z]=d%n;
			d/=n;
			z++;
		}
		d=0;
		long long k1=1;
		int f3=0;
		for(int i=z-1;i>=0;i--)
		{
			if(a[i]>0)f3=1;
			if(f3==0)continue;
			d+=k1*a[i];
			k1*=n;
		}
		for(long long i=2;i<=sqrt(d);i++)
		{
			if(d%i==0)
			f2=1;
		}
		if(d==1)
		f2=1;
		if(f1||f2)
		{
			cout<<"No"<<endl;
		}
		else cout<<"Yes"<<endl;
	}
	return 0;
}