PAT甲级 -- 1015 Reversible Primes (20 分)

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最新推荐文章于 2025-01-20 21:57:36 发布

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博客围绕可逆素数展开，给定两个正整数N和D，需判断N在基数D下是否为可逆素数。介绍了输入输出规范及示例，还提到解题时因英文题目易曲解题意，过程一波三折。

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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

AC:

一波三折的代码，最开始没搞懂题目，兜兜转转绕了很多圈.....

英文题目就是容易曲解题意哭

#include <iostream>
#include <string>
#include <algorithm>
#include <sstream>
#include <cmath>
using namespace std;

int N, D;
long long change = 0;
bool is_prime(long long num)
{
	if(num == 1 || num == 0) return false;
	long long sqr = sqrt(num*1.0);
	for(int i = 2; i <= sqr; i++)
	{
		if(num % i == 0) return false;
	}
	return true;
}

long long m2n(long long m, long long d)  //m转成进制
{
	long long arr[20] = {0}, index = 0;
	long long result = 0;
	while(m != 0)
	{
		arr[index++] = m % d;
		m /= d;
	}
	for(int i = 0; i < index; i++)
	{
		result = result * 10 + arr[i];
	}
	return result;
}

long long m2n10(long long m,long long d)  //m(d进制)转成10进制
{
	long long arr[20] = {0}, index = 0;
	long long result = 0;
	while(m != 0)
	{
		result += m % 10 * pow(d,index++);
		m /= 10;
	}
	return result;
}
int main()
{
	while(scanf("%d", &N) && N >= 0)
	{
		scanf("%d", &D);
		if(is_prime(N))
		{
			change = m2n(N,D);
			if(is_prime(m2n10(change,D)))
			{
				printf("Yes\n");
			}else
			{
				printf("No\n");
			}
		}else
		{
			printf("No\n");
		}

	}
	return 0;
}