IT panel ali end == main cout 取石子 bottom

取石子遊戲

Problem Description 1堆石子有n個,兩人輪流取.先取者第1次可以取任意多個，但不能全部取完.以後每次取的石子數不能超過上次取子數的2倍。取完者勝.先取者負輸出"Second win".先取者勝輸出"First win". Input 輸入有多組.每組第1行是2<=n<2^31. n=0退出. Output 先取者負輸出"Second win". 先取者勝輸出"First win".
參看Sample Output. Sample Input 2 13 10000 0 Sample Output Second win Second win First win (適用於特殊的題型)

 1
 
 #include<iostream>
 2 #include<map>
 3 using namespace std;
 4 int main()
 5 {
 6     int fib[500];
 7     fib[0] = 1, fib[1] = 2;
 8     for (int i = 2; i <= 70;i++)
 9         fib[i] = fib[i - 1] + fib[i - 2];
10     int n;
11     while (cin >> n&&n)
12     {
13         int
 
 mark = 0;
14         for (int i = 0; i <= 70; i++) {
15             if (fib[i] == n) mark = 1, cout << "Second win" << endl;
16             if (fib[i] > n) break;
17         }
18         if (!mark) cout << "First win" << endl;
19     }
20 
21     return 0;
22 }

斐波拉契博弈