1015. Reversible Primes (20)
时间限制
400 ms
内存限制
65536 kB
代码长度限制
16000 B
判题程序
Standard
作者
CHEN, Yue
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 1023 223 10-2
Sample Output:
YesYesNo
一个数N(为十进制)，判断他是否为质数
将N转换为D进制   翻转看其十进制的值是否为质数
如果都为质数Yes，否则No
例如：7 3  （7本身为质数；三进制表示21，那么翻转12，12的十进制值4，不是质数；综上结果No）
此题sqrt 求平方根函数，在头文件#include<math.h> 

评测结果
时间结果得分题目语言用时(ms)内存(kB)用户7月27日 09:39答案正确201015C++ (g++ 4.7.2)1436datrilla
测试点
测试点结果用时(ms)内存(kB)得分/满分0答案正确143612/121答案正确14362/22答案正确14364/43答案正确14362/2

#include<iostream>  #include<math.h>using namespace std;  int reverseN(int p, int radix){  int reverse = 0,i=0;  while (p>0)  {    reverse = reverse*radix + p % radix;    p /= radix;  }  return reverse;}bool PrImes(int p){  int i ;  if (p < 2)return false;  for (i = 2; i <= sqrt(p); i++)  {    if (0 == p%i)return false;  }  return true;}int main(){     int D, N;   while (cin >> N&&N>0)  {      cin >> D;    if (PrImes(N) && PrImes(reverseN(N, D)))cout << "Yes" << endl;    else cout << "No" << endl;  }   system("pause");    return 0;}