PAT Advanced 1015 Reversible Primes (20) [素数]

题目

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

题目分析

可翻转质数：十进制中的质数，转换为指定进制并反转后的数字仍然是质数
已知十进制数，和指定进制，判断是否为可翻转质数

解题思路

1. 判断输入的十进制数是否为质数
2. 将输入的十进制数转换为指定进制数A
3. 反转A得到B，判断B是否为质数

易错点

1. 题目已知条件中数字为十进制，而不是指定进制下的数，虽然题目中未说明，但是可以从样例中23不可能是2进制判断出

知识点

1. 题目没有给出样例数，输入非负数终止，该情况下的接收代码

while(scanf("%d",&n)!=EOF){
    if(n<0)break;
}

Code

Code 01

#include <iostream>
#include <algorithm>
#include <cmath>
using namespace std;
bool isPrime(int n) {
	if(n<=1)return false;
	int sqr=(int)sqrt(1.0*n);
	for(int i=2; i<=sqr; i++) {
		if(n%i==0)return false;
	}
	return true;
}
int main(int argc, char * argv[]) {
	int n,radix;
	while(scanf("%d",&n)!=EOF) {
		if(n<0)break;
		scanf("%d",&radix);
		if(isPrime(n)==false){
			printf("No
");
			continue;
		}
		// 进制转换
		// 十进制数转换为指定的radix进制 
		int index=0,d[111]={0};
		do{
			d[index++]=n%radix;
			n/=radix;
		} while(n!=0);
		// 逆置的radix进制数转换为十进制数 
		for(int i=0;i<index;i++){
			n=n*radix+d[i];
		}
		if(isPrime(n))printf("Yes
");
		else printf("No
");
	}
	return 0;
}