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 (<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

 

 题解：

一开始想直接用库函数做进制转换，结果itor不是标准库里的，只好自己写了。

算法笔记第二次转成十进制的算法是： n = n*radix + d[i]; 也可以，相当于从高位到低位算加法。

有懂原理的小伙伴麻烦评论里指点一下！！！

#include <iostream>
#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 d[111];
int main() {
	int n, radix;
	while (scanf("%d",&n) != EOF)
	{
		if (n < 0) break;
		scanf("%d", &radix);
		if (isPrime(n) == false) printf("No\n");
		else
		{
			int len = 0;
			do
			{
				d[len++] = n % radix;
				n /= radix;
			} while (n!=0);
			for (int i = 0; i < len; i++)
				n += d[i] * pow(radix, len - i - 1);
			if(isPrime(n) == true) printf("Yes\n");
			else printf("No\n");
		}
	}
    return 0;
}