1015 Reversible Primes (20)（20 分）

Posted 2020-11-10 巴蜀小小生

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

题目大意：给两个数N,d。判断N用d进制表示的反转数是否为质数，以及N本身是不是质数； 总的来说就是判断质数，进制转换
注意输入结束不是一-2为标志，而是以负数为标志。之前一直用-2作为结束的标志，浪费了不少的时间找bug

 1 #include<iostream>
 2 #include<cmath>
 3 using namespace std;
 4 
 5 bool is_prime(int num){
 6   if(num<2) return false;
 7   int temp = sqrt(num) + 1;
 8   for(int i=2; i<temp; i++)
 9     if(num%i==0) return false;
10   return true;
11 }
12 
13 int reverse(int num, int d){
14   int ans = 0, temp;
15   while(num){
16     temp = num%d;
17     ans = ans*d + temp;
18     num /= d;
19   }
20   return ans;
21 }
22 int main(){
23   int num, d;
24   while(true){
25     scanf("%d", &num);
26     if(num<0) break;
27     scanf("%d", &d);
28     if(is_prime(num) && is_prime(reverse(num, d))) printf("Yes\n");
29     else printf("No\n");
30   }
31   return 0;
32 }