[PAT 甲级] 1015 Reversible Primes （20 分) [进制转换]

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1015 Reversible Primes （20 分)

原题链接

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

---

题目的意思是，先判断n是不是一个素数，同时将n变成d进制的数，然后将这个数颠倒，然后将这个数变成10进制数，在判断这个数是否是一个素数。

我的方法是：先用取对数（换底公式）的方法先确定转化成d进制的位数，这样在转化的同时我们也就可以求除颠倒以后的10进制数，说起来比较难懂，直接看代码估计更容易理解。

int to_d(int n, int r) {
   int cnt = log(n)/log(r); //利用换底公式来求位数
   int num = 0;
   for (int i = cnt; i >= 0; i--) {
      num += (n % r) * pow(r,i);
      n /= r;
   }
   return num;
}

---

#include<iostream>
#include<string>
#include<cmath>
using namespace std;

int book[100001];
void getprime(int n) {
   book[1] = book[0] = 1;
   for (int i = 2; i <= n;i ++) {
      if (book[i] == 0) {
         for (int j = i + i; j <= n; j += i) {
            book[j] = 1;
         }
      }
   }
}

int to_d(int n, int r) {
   int cnt = log(n)/log(r);
   int num = 0;
   for (int i = cnt; i >= 0; i--) {
      num += (n % r) * pow(r,i);
      n /= r;
   }
   return num;
}
int main () {
   getprime(100001);
   int n, d;
   while(true) {
      cin >> n;
      if (n < 0) {
         break;
      } else {
         cin >> d;
      }
      int tem = to_d(n, d);
      if (book[n] == 0 && book[tem] == 0) {
         cout << "Yes" << '\n';
      } else {
         cout << "No" << '\n';
      }
   }
   return 0;
}