斐波拉契博弈

取石子游戏

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.

 

Sample Input

2 13 10000 0

 

Sample Output

Second win Second win First win

(适用于特殊的题型)

 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 }

 

来源：https://www.cnblogs.com/kangdong/p/9190901.html