1015 Reversible Primes (20分)（一个字水）

1015 Reversible Primes (20分)

分析

打印素数表，判断给定的数以及其在给定进制下的转置数是否为素数

代码

#include<iostream>
using namespace std;
int n, d, nonprime[100010];
int getreverse(int n,int d) {
 int s = 0;
 while (n != 0) {
  s = s * d + n % d;
  n /= d;
 }
 return s;
}
int main() {
 for (int i = 2; i * i < 100010; ++i)
  for (int j = 2; j * i < 100010; ++j)
   nonprime[i * j] = 1;
 nonprime[0] = nonprime[1] = 1;
 while (true) {
  scanf("%d", &n);
  if (n < 0)break;
  scanf("%d", &d);
  if (nonprime[n] || nonprime[getreverse(n, d)])printf("No\n");
  else printf("Yes\n");
 }
 return 0;
}