1015. Reversible Primes (20)

最新推荐文章于 2022-02-22 19:31:52 发布

empty_coder

最新推荐文章于 2022-02-22 19:31:52 发布

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分类专栏： PAT PAT 文章标签： pat

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题目链接:https://www.patest.cn/contests/pat-a-practise/1015

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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

分析：判断n以及rn(10进制n转化为d进制dn,然后反转为rdn,最后转化为10进制rn)是否为质数

吐槽：PAT的一些题目。。。理解题意可比算法/代码实现困难多啦，或者我太过愚钝？

#include <stdio.h>
#include <math.h>

int n,rn,d;

int Prime(int N){
	if(N == 1) return 0; //单独处理 
	for(int i=2;i<=(int)sqrt(N);i++){
		if(N%i == 0) return 0;
	}
	return 1;
}

void CRC(){ 
	int tmp = n;
	rn = 0;
	while(tmp){
		rn = rn*d+tmp%d;	
		tmp /= d;
	}
}
 
int main(){
	while(1){
		scanf("%d",&n);
		if(n < 0) break;
		scanf("%d",&d);
		if(Prime(n)){
			CRC();
			if(Prime(rn)) printf("Yes\n");
			else printf("No\n"); 
		}
		else printf("No\n");
	}
	return 0;
}

分析部分参照： http://xujiayu317.blog.163.com/blog/static/254752092015627946382