PAT甲级——A1015 Reversible Primes
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 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
其实就是将输入的数字转化为该进制的反序列,然后再转化为10进制
然后判断是否是素数即可
1 #include <iostream> 2 #include <math.h> 3 #include <string> 4 5 using namespace std; 6 7 bool isPrim(int a) 8 { 9 if (a < 2) 10 return false; 11 int b = (int)sqrt(1.0 * a); 12 for (int i = 2; i <= b; ++i) 13 { 14 if (a%i == 0) 15 return false; 16 } 17 return true; 18 } 19 20 int getReverNum(int a, int b) 21 { 22 string str = "";//得到反转的序列 23 while (a > 0) 24 { 25 str += a % b + '0'; 26 a /= b; 27 } 28 int n = 0; 29 for (int i = str.length() - 1, j = 0; i >= 0; --i, ++j) 30 n += (str[i] - '0')*pow(b, j); 31 return n; 32 } 33 34 int main() 35 { 36 int a, b; 37 while (cin >> a) 38 { 39 if (a < 0)break; 40 cin >> b; 41 if (isPrim(a) && isPrim(getReverNum(a,b))) 42 cout << "Yes" << endl; 43 else 44 cout << "No" << endl; 45 } 46 return 0; 47 }