HDU 3951 Coin Game（博弈）

Coin Game

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2000    Accepted Submission(s): 1119

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

 

Author

NotOnlySuccess

 

Source

2011 Alibaba Programming Contest

 

Recommend

lcy

  最优策略：当 k > 1 时，后手只需拿走先手对面的一到两枚硬币（如图），使左右两边硬币数相等，则必胜。

#include<bits/stdc++.h>using namespace std;int main(){    int t,p=1;    scanf("%d",&t);    while(t--)    {        int n,k;        scanf("%d%d",&n,&k);        printf("Case %d: ",p++);        if(k==1)        {            if(n&1) printf("first\n");            else printf("second\n");        }        else if(k>=n)            printf("first\n");        else            printf("second\n");    }}