取石子游戏

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 8159    Accepted Submission(s): 4950

Problem Description1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. Sample Input
2
13
10000
0
 Sample Output
Second win
Second win
First win

C/C++：

 #include <map>
 #include <queue>
 #include <cmath>
 #include <vector>
 #include <string>
 #include <cstdio>
 #include <cstring>
 #include <climits>
 #include <iostream>
 #include <algorithm>
 #define INF 0x3f3f3f3f
 #define LL long long
 #define wzf ((1 + sqrt(5.0)) / 2.0)
 using namespace std; __int64 n, fib[] = {, }; void calc()
 {
     for (__int64 i = ; i <= ; ++ i)
         fib[i] = fib[i - ] + fib[i - ];
 } bool is_fib()
 {
     for (__int64 i = ; i <= ; ++ i)
     {
         if (fib[i] > n) return false;
         if (n == fib[i])
             return true;
     }
     return false;
 } int main()
 {
     calc();
     while (scanf("%I64d", &n), n)
     {
         if (is_fib())
             printf("Second win\n");
         else
             printf("First win\n");
     }
     return ;
 }

C/C++:

 #include <map>
 #include <set>
 #include <queue>
 #include <cmath>
 #include <vector>
 #include <string>
 #include <cstdio>
 #include <cstring>
 #include <climits>
 #include <iostream>
 #include <algorithm>
 #define INF 0x3f3f3f3f
 #define LL long long
 using namespace std; __int64 n, num[] = {, }; set <__int64> s; void calc()
 {
     s.insert();
     for (int i = ; i <= ; ++ i)
     {
         num[i] = num[i - ] + num[i - ];
         s.insert(num[i]);
     }
 } int main()
 {
     calc();
     while (scanf("%I64d", &n), n)
     {
         if (s.find(n) != s.end())
             printf("Second win\n");
         else
             printf("First win\n");
     }
     return ;
 }