pat 1015 Reversible Primes(20 分)
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1015 Reversible Primes(20 分)
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
1 #include <map> 2 #include <set> 3 #include <queue> 4 #include <cmath> 5 #include <stack> 6 #include <vector> 7 #include <string> 8 #include <cstdio> 9 #include <cstring> 10 #include <climits> 11 #include <iostream> 12 #include <algorithm> 13 #define wzf ((1 + sqrt(5.0)) / 2.0) 14 #define INF 0x3f3f3f3f 15 #define LL long long 16 using namespace std; 17 18 const int MAXN = 1e4 + 10; 19 20 int n, r; 21 22 bool is_prime(int n) 23 { 24 if (n == 0 || n == 1) return false; 25 for (int i = 2; i * i <= n; ++ i) 26 if (n % i == 0) return false; 27 return true; 28 } 29 30 int main() 31 { 32 while (scanf("%d", &n), n >= 0) 33 { 34 scanf("%d", &r); 35 if (!is_prime(n)) 36 { 37 printf("No "); 38 continue; 39 } 40 int cnt = 0, ans = 0; 41 stack <int> my_stack; 42 while (n) 43 { 44 my_stack.push(n % r); 45 n /= r; 46 } 47 while (my_stack.size()) 48 { 49 ans += my_stack.top() * (pow(r, cnt)); 50 ++ cnt; 51 my_stack.pop(); 52 } 53 if (!is_prime(ans)) 54 printf("No "); 55 else 56 printf("Yes "); 57 } 58 return 0; 59 }
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