（k倍动态减法游戏）zoj 3599 hdu 2486

A simple stone game

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 359    Accepted Submission(s): 190

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

  

   5
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
  

 

题解

http://blog.youkuaiyun.com/acm_cxlove/article/details/7836544

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int maxn=4000000+1000;
int a[maxn];
int b[maxn];
int main()
{
    int ca=1,t;
    scanf("%d",&t);
    while(t--)
    {
        int n,k;
        scanf("%d%d",&n,&k);
        a[0]=b[0]=1;
        int i,j;
        i=j=0;
        while(b[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]=a[i]+b[j];
            else
            b[i]=a[i];
        }
        printf("Case %d: ",ca++);
        if(a[i]==n)
        printf("lose\n");
        else
        {
            int ans;
            while(n)
            {
                if(a[i]<=n)
                {
                    n-=a[i];
                    ans=a[i];
                }
                i--;
            }
            printf("%d\n",ans);
        }
    }
    return 0;
}

zoj2599

Game

---

Time Limit: 3 Seconds       Memory Limit: 65536 KB

---

Alice and Bod decide to play a new stone game. At the beginning Alice puts n(n>1) stones (out of N in all) on the table. Alice and Bob remove the stones in turn. At each step the player should remove some number of stones. The number of stones the player removed should be at least one, and cannot exceed m times of the number of stones the player removed at the last step. The player who removes the last stone wins the game. Alice always plays first and of course at the first turn, she cannot remove all the stones. Alice wants to know how many positive integers n she can choose to win the game if both players play optimally.

Since you're an ace programmer, Alice wants you to help her.

Input

There are multiple test cases. The first line of input contains an integer T (0 < T ≤ 500) indicating the number of test cases. Then T test cases follow.

Each test case is a line of 2 integers m and N(0 < m ≤ 2012, 1 < N < 2³¹)

Output

For each test case output the number of positive integers Alice can choose for n to make her win the game.

Sample Input

3
1 10
2 10
3 10

Sample Output

6
5
4

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
const int maxn=2000000+100;
long long a[maxn],b[maxn];
using namespace std;

int main()
{
    long long n,k;
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%lld%lld",&k,&n);
        int i,j;
        i=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];
        }
        long long ans;
        if(a[i]==n)
        ans=n-i-1;
        else
        ans=n-i;
        printf("%lld\n",ans);

    }
    return 0;
}