Hdu 3951 Coin Game【对称博弈】

Coin Game

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1997    Accepted Submission(s): 1117

Problem 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<=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

23 13 2

 

Sample Output

Case 1: firstCase 2: second

题目大意：

给你排成一圈N个硬币，每次可以连续拿相邻位子上的K个硬币出来，问先手赢还是后手赢。

赢的玩家是拿走最后一枚硬币的玩家。

思路：

对称性博弈，如果N%2==0.那么一定是后手赢，因为要将硬币排成一个圆，如果先手有的拿，那么后手对称着到圆的另一侧去拿即可。

如果N%2==1.如果此时k==1.那么一定是先手赢。

如果此时k>1，那么同理也一定可以构成后手对称的情况，所以后手一定也是赢的。

Ac代码：

#include<stdio.h>#include<string.h>using namespace std;int main(){    int t;    int kase=0;    scanf("%d",&t);    while(t--)    {        int n,k;        scanf("%d%d",&n,&k);        printf("Case %d: ",++kase);        if(k>=n)        {            printf("first\n");            continue;        }        if(n%2==0)        {            printf("second\n");        }        else        {            if(k==1)printf("first\n");            else printf("second\n");        }    }}