HDU 3951 Coin Game(博弈取对称思路)

Coin Game

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1494 Accepted Submission(s): 853

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 

/*（转分析：） 题意：有N个硬币围成一圈，两个人每次只能取连续的1~K个硬币。 问先手有没有必胜策略。解法：当K==1：结果只与N的奇偶有关。     当K>=2：如果N<=K，那么很显然先手必胜。             当N>K时，先手一下子取不完，设剩余M个，   如果M<=K则后手一下全部取完，后手必胜；当M>K，即此时M>=3且K>=2，那么后手就可以在剩下的一排里取中间留两边然后以后就对称取，就一定必胜；所有此时后手必胜；*/#include <cstdio>#include <iostream>using namespace std;int main(){int t,i,n,k,now;scanf("%d",&t);       for(now=1;now<=t;now++){ scanf("%d%d",&n,&k);printf("Case %d: ",now);if(k==1){if(n%2==1)printf("first\n");elseprintf("second\n");}else{          if(k>=n)printf("first\n"); elseprintf("second\n");}}return 0;}