取石子游戏
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3605 Accepted Submission(s): 2127
Total Submission(s): 3605 Accepted Submission(s): 2127
Problem Description
1堆石子有n个,两人轮流取.先取者第1次可以取任意多个,但不能全部取完.以后每次取的石子数不能超过上次取子数的2倍。取完者胜.先取者负输出"Second win".先取者胜输出"First win".
Input
输入有多组.每组第1行是2<=n<2^31. n=0退出.
Output
先取者负输出"Second win". 先取者胜输出"First win".
参看Sample Output.
参看Sample Output.
Sample Input
2 13 10000 0
Sample Output
Second win Second win First win
Source
判断是不是斐波数就好
#include<stdio.h> #include<string.h> int f[50]; void fib() { f[1]=f[2]=1; for(int i=3;i<50;i++) f[i]=f[i-1]+f[i-2]; } int main() { int n; while(scanf("%d",&n),n) { fib(); int flog=0; for(int i=0;i<50;i++) { if(f[i]==n) { flog=1; break; } } if(flog) printf("Second win "); else printf("First win "); } return 0; }