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 (<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

题目比较水，但是第二个点一直没过，参考别人的博客，才突然发现，1不是素数。

代码

#include<cstdio>
#include<cmath>
#include<cstring>

using namespace std;

const int maxn = 1e6 + 100;
int a[maxn],sum;

void solve(int n, int radix){
  memset(a, 0, sizeof(a));
  sum = 0;
  int res = 0;
  int t = n, tem;
  while(t){
    tem = t%radix;
    t /= radix;
    a[sum++] = tem;
  }
  bool flag = false;
  int bei = 1;
  for(int i = sum-1; i >= 0; i--) {
    res += bei*a[i];
    bei *= radix;
  }
  if(res == 1){
    printf("No\n");
    return;
  }
  for(int i = 2; i <= sqrt(res); i++) {
    if(res % i == 0) {
      printf("No\n");
      return;
    }
  }
  printf("Yes\n");
}

int main(){
  int n,d;
  while(~scanf("%d",&n) && n >= 0){
    scanf("%d",&d);
    bool f = false;
    if(n == 1)f = true;
    for(int i = 2; i <= sqrt(n); i++) {
      if(n % i == 0) {
        f = true;
        break;
      }
    }
    if(f){
      printf("No\n");
      continue;
    }
    solve(n,d); 
  }
  return 0;
}