博弈论

　　　　　　　　　　　　　　　　Coin Game

After hh has learned how to play Nim game, he begins to try another coin game which seems much easier.

The game goes like this:
Two players start the game with a circle of n coins.
They take coins from the circle in turn and every time they could take 1~K continuous coins.
(imagining that ten coins numbered from 1 to 10 and K equal to 3, since 1 and 10 are continuous, you could take away the continuous 10 , 1 , 2 , but if 2 was taken away, you couldn't take 1, 3, 4, because 1 and 3 aren't continuous)
The player who takes the last coin wins the game.
Suppose that those two players always take the best moves and never make mistakes.
Your job is to find out who will definitely win the game.

 

Input

The first line is a number T(1<=T<=100), represents the number of case. The next T blocks follow each indicates a case.
Each case contains two integers N(3<=N<=10⁹,1<=K<=10).

 

Output

For each case, output the number of case and the winner "first" or "second".(as shown in the sample output)

 

Sample Input

2

3 1

3 2

 

Sample Output

Case 1: first

Case 2: second

 

题目大意： 将硬币1到硬币n呈圆形摆放，首尾相接，（如上图），A，B两人轮流取一些枚硬币，硬币的数量不超过K并且必须连续，（连续即所取硬币编号连续）谁取走最后一枚硬币谁就获胜；T组样例，每组样例输入n,k,输出获胜者。

 

分析：博弈论的题，师哥说是对称博弈，B必胜的套路是将所有的硬币分成对称的两部分，之后A怎么取，B就怎么取，这样B一定能取到最后一枚硬币。

具体到此题：

三种情况：

1. A第一次全部取完 A胜
2. B不能将所有的硬币分成对称的两部分。
3. B完成套路。

代码实现：

#include <stdio.h>
int main()
{
    int e,i;
    scanf("%d",&e);
    for(i=1;i<=e;i++)
    {
        int n,k,d=0;
        scanf("%d%d",&n,&k);
        if(n<=k)
            d=1;
        else
        {
            if(n==1|| (n==3&&k==1) || (n>3&&n%2!=0&&k==1) )
                d=1;
            else
                d=2;
        }
        if(d==1)
            printf("Case %d: first
",i);
        else
            printf("Case %d: second
",i);
    }
    return 0;
}