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
思路
先判断n是不是素数，然后将n转化为d进制的数，在把这个数翻转，然后在从d进制转化为10进制数，然后在判断是不是素数。
Code
#include <iostream>
#include <math.h>
using namespace std;
// 将n转化为d进制(1<=d<=10)的数
// 将d进制的数进行翻转
string convert_to_d_radix(int n, int radix)
{
// 10进制转换为d进制
string temp = "";
while (n != 0)
{
int mod = n % radix;
temp += to_string(mod);
n = n / radix;
}
return temp;
}
// 将翻转得到的数转换为10机制
int convert(string n, int radix)
{
//d进制转换为10进制
int sum = 0;
int index = 0;
for (int i = n.size() - 1; i >= 0; i--)
{
int temp = n[i] - '0';
sum += temp * pow(radix, index++);
}
return sum;
}
// 判断10进制的数是不是素数
bool is_prime(int n)
{
if (n == 1)
return false; // 1不是素数
for (int i = 2; i <= sqrt(n); i++)
{
if (n % i == 0)
{
return false; // 不是素数
}
}
return true;
}
int main()
{
int d;
int n;
cin >> n;
while (n >= 0)
{
cin >> d;
if (is_prime(n))
{
if (is_prime(convert(convert_to_d_radix(n, d), d)))
{
cout << "Yes" << endl;
}
else
{
cout << "No" << endl;
}
}
else
{
cout << "No" << endl;
}
cin >> n;
}
return 0;
}