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 (<105) 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
Experiential Summing-up
判断输入是否为负数，判断n是否为素数，把n转换为d进制再反过来转换为10进制，判断是否为素数。
例如：23是素数，转换成2进制为11101，反转后为10111，转换成10进制数为29，还是素数，输出Yes
Accepted Code
1 #include<bits/stdc++.h> 2 using namespace std; 3 4 bool isprime(int n) 5 { 6 if(n <= 1) return false; 7 for(int i = 2; i <= n/2; i ++) 8 if(n % i == 0) 9 return false; 10 return true; 11 } 12 13 int main() 14 { 15 int n, d; 16 while(scanf("%d", &n) != EOF) 17 { 18 if(n < 0) break; 19 cin >> d; 20 if(isprime(n) == false) 21 { 22 cout << "No" << endl; 23 continue; 24 } 25 int len = 0, arr[100]; 26 do 27 { 28 arr[len++] = n % d; 29 n = n / d; 30 } while(n != 0); 31 for(int i = 0; i < len; i ++) 32 n = n * d + arr[i]; 33 printf("%s", isprime(n) ? "Yes\n" : "No\n"); 34 } 35 return 0; 36 }