PAT_A1015#Reversible Primes

Source:

PAT A1015 Reversible Primes (20 分)

Description:

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 (<) and D (1), 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 Nand 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

Keys:

• 进制转换
• 素数（Prime）

Attention:

• int < 1e9 ~ 2^8-1, long < 1e18 ~ 2^16-1；

• 1e5的十进制转换为二进制会超过int范围，进制转换和逆置一起进行的话就避免中间数超出范围的问题了；

Code:

 /*
 Data: 2019-05-12 21:32:05
 Problem: PAT_A1015#Reversible Primes
 AC: 18:26

 题目大意：
 给一整数N<1e5，进制D
 求N的D进制逆置后是否仍是素数
 */
 #include<cstdio>
 #include<string>
 #include<algorithm>
 using namespace std;
 const int M=1e5+;

 bool isPrime(int num)
 {
     if(num== || num==)
         return false;
     for(int i=; i*i<=num; i++)
         if(num%i==)
             return false;
     return true;
 }

 int Reverse(int n, int d)
 {
     int ans=;
     while(n!=)
     {
         ans *= d;
         ans += (n%d);
         n /= d;
     }
     return ans;
 }

 int main()
 {
 #ifdef  ONLINE_JUDGE
 #else
     freopen("Test.txt", "r", stdin);
 #endif // ONLINE_JUDGE

     int n,d;
     while(scanf("%d",&n) && n>=)
     {
         scanf("%d", &d);
         if(isPrime(n) && isPrime(Reverse(n,d)))
             printf("Yes\n");
         else
             printf("No\n");
     }

     return ;
 }