FAFU OJ 博弈(10)乘法遊戲
阿新 • • 發佈:2018-12-23
博弈(10)乘法遊戲
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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 17 34012226 Sample Output: Z wins. C wins. Z wins. Source:
Time Limit: | 1000MS | Memory Limit: | 65536KB |
Total Submissions: | 46 | Accepted: | 28 |
Z wins.
or
C wins.
assuming that both of them play perfectly.
Sample Input: 162 17 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; }