hdu3951Coin Game(对称博弈)

时间：2018-10-23 18:10:58 阅读：134 评论：0 收藏：0 [点我收藏+]

标签：else 博弈 contain 技术分享 start cond col contains java

Coin Game

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2947 Accepted Submission(s): 1609

Problem Description

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

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

3 1

3 2

Sample Output

Case 1: first

Case 2: second

题意：硬币围成一圈，编号1-n，每人每次最多取k个连续的硬币，给出n和k，问最后谁赢

题解：典型的对称博弈，当k==1时，就看n的奇偶，k>=2时，应用对称博弈：先手取完后，环变成了链，后手只要创建一个对称局势便可获得胜利，把这条链分成相等的链

k>=2时，后手一定可以创建出对称局势，此时后手必胜

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 int main() {
 4     int t;
 5     while(~scanf("%d",&t))
 6     {
 7         
 8         for(int i=1;i<=t;i++)
 9         {
10             int n,k;
11             scanf("%d %d",&n,&k);
12             if(k>=n)
13             {
14                 printf("Case %d: first\n",i);
15             }else
16             {
17                 if(k==1)
18                 {
19                     if(n%2==0)printf("Case %d: second\n",i);
20                     else printf("Case %d: first\n",i);
21                 }else
22                 {
23                     printf("Case %d: first\n",i);
24                 }
25             }
26         }
27     }
28     return 0;
29 }

hdu3951Coin Game(对称博弈)

标签：else 博弈 contain 技术分享 start cond col contains java

原文地址：https://www.cnblogs.com/fqfzs/p/9837608.html

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