Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6590 Accepted Submission(s):
3981
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
Recommend
裸的斐波那契博弈
当输入的数据为斐波那契数的时候先手胜，否则后手胜
#include<cstdio> #include<map> int fib[233],x; std::map<int,bool>mp; int main() { fib[1]=1;fib[2]=1; for(int i=3;i<=50;i++) fib[i]=fib[i-1]+fib[i-2],mp[fib[i]]=1; while(scanf("%d",&x)&&x!=0) puts(mp[x]==1?"Second win":"First win"); return 0; }