A simple stone game

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 815 Accepted Submission(s): 474

Problem Description

After he has learned how to play Nim game, Mike begins to try another stone game which seems much easier.

The game goes like this: Two players start the game with a pile of n stones. They take stones from the pile in turn and every time they take at least one stone. The one who goes first can take at most n-1 stones for his first move. From then on a player can take at most k times as many stones as his opponent has taken last time. For example, if one player take m stones in his turn, then the other player can take at most k × m stones next time. The player who takes the last stone 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 contains a integer t, indicating that there are t test cases following.(t<=20).

Each test case is a line consisting of two integer n and k.(2<=n<=10^8,1<=k<=10^5).

Output

For each test case, output one line starting with “Case N: ”, N is the case number. And then, if the first player can ensure a winning, print the minimum number of stones he should take in his first turn. Otherwise, print "lose". Please note that there is a blank following the colon.

Sample Input

16 1

11 1

32 2

34 2

19 3

Sample Output

Case 1: lose

Case 2: 1

Case 3: 3

Case 4: lose

Case 5: 4

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<queue>
#include<math.h>
#include<map>
#include<vector>
#include<stack>
#include<iostream>
#define inf 0x3f3f3f3f
using namespace std;
typedef long long ll;
const int N=20000005;

int a[N],b[N];

int main()
{
    int n,k,T,cas=0;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d %d",&n,&k);
        int i=0,j=0;
        a[0]=b[0]=1;
        while(a[i]<n)
        {
            i++;
            a[i]=b[i-1]+1;
            while(a[j+1]*k<a[i])
                j++;
            if(a[j]*k<a[i])
                b[i]=b[j]+a[i];
            else
                b[i]=a[i];
        }
        printf("Case %d: ",++cas);
        if(n==a[i])
            printf("lose\n");
        else
        {
            int ans;
            while(n)
            {
                if(n>=a[i])
                {
                    n-=a[i];
                    ans=a[i];
                }
                i--;
            }
            printf("%d\n",ans);
        }
    }
    return 0;
}

转自 http://blog.csdn.net/ACM_cxlove?viewmode=contents

神牛曹钦翔的论文：从“k倍动态减法游戏”出发探究一类组合游戏问题

反正偶是没看懂。。。。

两人取一堆n个石子先手不能全部取完之后每人取的个数不能超过另一个人上轮取的数*K

给n，K判断先手必胜并求第一步

博弈题

这题的思考过程非常有意义。

当k=1的时候可知必败局面都是2^i 将n分解成二进制，然后先手取掉最后一个1.然后对方必然无法去掉更高的1，而对方取完我方至少还能拿掉最后一个1 导致对方永远取不完。

当k=2的时候，必败局面都是斐波那契数列。利用“先手去掉最后一个1，则后手必不能去掉更高阶的1导致取不完”的思想，斐波那契数列有一个非常好的性质就是：任意一个整数可以写成斐波那契数列中的不相邻的项的和，于是将n写成这种形式，先取走最后一个1，对方能取的数是这个数*2，小于高2位的1，所以取不完。

当K的时候，想办法构造数列，将n写成数列中一些项的和，使得这些被取到的项的相邻两个倍数差距>k 那么每次去掉最后一个1 还是符合上面的条件。设这个数列已经被构造了i 项，第 i 项为a[ i ]，前 i 项可以完美对1..b[ i ] 编码使得每个编码的任意两项倍数>K 那么有

a[ i+1 ] = b[ i ] + 1;这是显然的因为b[ i ] + 1没法构造出来，只能新建一项表示

然后计算b[ i+1] 既然要使用 a[ i+1 ] 那么下一项最多只能是某个 a[ t ] 使得 a[ t ] * K < a[ i+1 ] 于是

b[ i ] = b[ t ] + a[ i+1 ]

然后判断n是否在这个数列里面

如果在，那么先手必败。否则不停的减掉数列a中的项构造出n的分解，最后一位就是了。

以上转自神牛：http://hi.baidu.com/lccycc_acm/item/a6f0dd0ec5c44a39f3eafcd3

做一些解释：首先是a[i]=b[i-1]+1; b[i-1]是由a[0]……a[i-1]组成的最大的数，那么b[i-1]+1不可能用a[0]……a[i-1]组成。

然后是：if(a[j]*k<a[i]) b[i]=b[j]+a[i]; else b[i]=a[i];

要求b[j]，表示a[0]……a[i]组成，那么显然是要用到a[i]的，不然不就成了b[i-1]，既然用了a[i]，但是又要使相邻的倍数在K以上。则找到最大的j，使

a[j]*k<a[i]那么满足条件，便是a[0]……a[j]能组成的最大的数，加上a[i]，那么后者表示当前项不能和之前项组合，那么最大的数就只能是本身

A simple stone game HDU 2486

A simple stone game

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