1015 Reversible Primes (20)

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 (< 10^5^) 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

可逆素数，就是指一个数是素数，它自身reverse之后仍是素数

给N（<10^5）和D（1

若是可逆素数则输出yes，否则no

C++：

#include "cstdio"
#include "iostream"
#include "cmath"
using namespace std;

//判断是否是素数
bool isPrime(int x){
	if (x<=1)
	{
		return false;
	}//要在这里做判断 否则有个测试点过不去 <=1的数都不是素数
	int sqr = int(sqrt(x*1.0));//开根数
	for (int i=2;i<=sqr;i++)
	{
		if (x%i==0)
		{
			return false;
		}
	}
	return true;
}

int main(){
	int n,d;
	int arr[100001];//储存非10进制的reverse
	bool flag=false;
	n=0;
	while (scanf("%d", &n) != EOF)
	{
		if (n<0)//小于0
		{
			break;
		}
		cin>>d;
		if (isPrime(n)==false)//本身就不是素数
		{
			cout<<"No"<=0;i--){
			reverse+=arr[i]*pow(d,j);//将reverse从arr中转化为数字
			j++;
		}
		flag=isPrime(reverse);
		if (flag)
		{
			cout<<"Yes"<