HDU 3951 Coin Game（博弈）

思路：一个环形的巴什博奕。如果k>=n显然就是先手必胜，而手推了一下发现，如果k为1的时候则看n的奇偶性，而k不为1的时候，无论先手取哪里，只要后手选择与之对称的点，最后一定是后手必胜。

#include<bits/stdc++.h>
using namespace std;

int main()
{
int T,cas=1;
scanf("%d",&T);
while(T--)
{
int n,k;
scanf("%d%d",&n,&k);
if(k>=n)
printf("Case %d: first\n",cas++);
else
{
if(k==1)
{
if(n&1)
printf("Case %d: first\n",cas++);
else
printf("Case %d: second\n",cas++);
}
else
printf("Case %d: second\n",cas++);
}
}
}


Description

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<=10 9,1<=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