FAFU OJ 博弈（10）乘法游戏 Z and C play the game of multiplicat

博弈（10）乘法游戏

Time Limit:	1000MS	Memory Limit:	65536KB
Total Submissions:	46	Accepted:	28

Description:

Z and C play the game of multiplication by multiplying an integer p by one of the numbers 2 to 9. Z always starts with p = 1, does his multiplication, then C multiplies the number, then Z and so on. Before a game starts, they draw an integer 1 < n < 4294967295 and the winner is who first reaches p >= n.

Input:

Each line of input contains one integer number n.

Output:

For each line of input output one line either

Z wins.

or

C wins.

assuming that both of them play perfectly.

Sample Input:

162

34012226

Sample Output:

Z wins.

C wins.

Z wins.

Source:

#include<iostream>
#include<cstdio>
using namespace std;
int main()
{
  double n;
  while(scanf("%lf",&n)!=EOF)
  {
     while(n>18) n/=18;
     if(n<=9)
     {
         printf("Z wins.\n");
     }
     else
     {
         printf("C wins.\n");
     }
  }
  return 0;
}