PAT A1015 Reversible Primes (20)

PAT A1015 Reversible Primes
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

分析：

1. 判断质数+进制转换：转换进制的时候用do...while...循环来保证0转换正确

#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>
using namespace std;
const int maxn = 100005;
bool vis[maxn];
vector<int> prime;
void euler(int n) {
  for (int i = 2; i <= n; i++) {
    if (!vis[i])
      prime.push_back(i);
    for (int j = 0; j < prime.size() && i * prime[j] <= n; j++) {
      vis[i * prime[j]] = true;
      if (i % prime[j] == 0)
        break;
    }
  }
}
bool IsPrime(int n) {
  vector<int>::iterator it = lower_bound(prime.begin(), prime.end(), n);
  if (it != prime.end() && *it == n)
    return true;
  else
    return false;
}
void DtoK(int n, vector<int> &nink, int k) {
  nink.clear();
  while (n != 0) {
    nink.push_back(n % k);
    n /= k;
  }
}
void KtoD(vector<int> nink, int &n, int k) {
  n = 0;
  int weigh = 1;
  for (int i = 0; i < nink.size(); i++) {
    n += nink[i] * weigh;
    weigh *= k;
  }
}
int main() {
  euler(maxn);
  int n, rev_n, d;
  vector<int> nink;
  while (scanf("%d", &n), n >= 0) {
    scanf("%d", &d);
    bool flag1=false, flag2(false);
    flag1 = IsPrime(n);
    DtoK(n, nink, d);
    reverse(nink.begin(), nink.end());
    KtoD(nink, rev_n, d);
    flag2 = IsPrime(rev_n);
    if (flag1 && flag2)
      printf("Yes\n");
    else
      printf("No\n");
  }
  return 0;
}

#include <cstdio>
#include <cmath>
using namespace std;
bool isprime(int n) {
    if(n <= 1) return false;
    int sqr = int(sqrt(n * 1.0));
    for(int i = 2; i <= sqr; i++) {
        if(n % i == 0)
            return false;
    }
    return true;
}
int main() {
    int n, d;
    while(scanf("%d", &n) != EOF) {
        if(n < 0) break;
        scanf("%d", &d);
        if(isprime(n) == false) {
            printf("No\n");
            continue;
        }
        int len = 0, arr[100];
        do{
            arr[len++] = n % d;
            n = n / d;
        }while(n != 0);
        for(int i = 0; i < len; i++)
            n = n * d + arr[i];
        printf("%s", isprime(n) ? "Yes\n" : "No\n");
    }
    return 0;
}