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
AC代码:(素数+进制转换(转化为十进制))
#include <iostream>
#include <bits/stdc++.h>
using namespace std;
int p[(int)1e7+10];
void prime()
{
p[0] = 1;
p[1] = 1;
for(int i=2; i<=(int)1e7+10; i++)
{
if(p[i]==0)
{
for(int j=i+i; j<=(int)1e7+10; j=j+i)
{
p[j] = 1;
}
}
}
}
int main()
{
prime();
int n,d;
queue<int>q;
while(~scanf("%d%d",&n,&d))
{
if(n<0)
break;
if(p[n]==0)
{
int m = n;
while(m)
{
q.push(m%d);
m = m / d;
}
int s = 0;
while(!q.empty())
{
s=s*d+q.front();
q.pop();
}
// printf("%d\n",s);
if(p[s]==0)
{
printf("Yes\n");
}
else
{
printf("No\n");
}
}
else
{
printf("No\n");
}
}
return 0;
}

本文探讨了可逆素数的概念,即在任何数制下,一个素数的反转也是素数的情况。通过给出具体的例子和算法实现,文章详细解释了如何判断一个数在特定进制下是否为可逆素数。
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