hdu 5795 A Simple Nim(2016 Multi-University Training Contest 6——博弈)

题目大意：http://acm.hdu.edu.cn/showproblem.php?pid=5795

A Simple Nim

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 550    Accepted Submission(s): 363

Problem Description

Two players take turns picking candies from n heaps,the player who picks the last one will win the game.On each turn they can pick any number of candies which come from the same heap(picking no candy is not allowed).To make the game more interesting,players can separate one heap into three smaller heaps(no empty heaps)instead of the picking operation.Please find out which player will win the game if each of them never make mistakes.

 

Input

Intput contains multiple test cases. The first line is an integer 1≤T≤100, the number of test cases. Each case begins with an integer n, indicating the number of the heaps, the next line contains N integers s[0],s[1],....,s[n−1], representing heaps with s[0],s[1],...,s[n−1] objects respectively.(1≤n≤106,1≤s[i]≤109)

 

Output

For each test case,output a line whick contains either"First player wins."or"Second player wins".

 

Sample Input

224 431 2 4

 

Sample Output

Second player wins.First player wins.

 

Author

UESTC

 

Source

2016 Multi-University Training Contest 6

 

题目大意：

两人于最优策略对n堆物品进行抓取，不能抓取的人输。

1、从同一堆中取任意个（>=0）

2、把一堆分成任意三堆（每一堆至少有一个）

解题思路：

sg[0]=0

当x=8k+7时sg[x]=8k+8，

当x=8k+8时sg[x]=8k+7,

其余时候sg[x]=x；（k>=0）

打表找规律可得，数学归纳法可证。

先附上打表代码。

#include <iostream>#include <cstdio>#include <cstring>using namespace std;int sg[500];int dfs(int n){    if(sg[n]!=-1)        return sg[n];    int vis[500];    memset(vis,0,sizeof(vis));    for(int i=0; i<n; i++)    {        vis[dfs(i)]=1;    }    for(int i=1; i<n; i++)    {        for(int j=1; j+i<n; j++)        {            vis[dfs(i)^dfs(j)^dfs(n-i-j)]=1;        }    }    int i;    for(i=0; i<500; i++)        if(vis[i]==0)            break;    return sg[n]=i;}int main (){    memset(sg,-1,sizeof(sg));    sg[0]=0;    for(int i=1; i<=50; i++)    {        sg[i]=dfs(i);        printf("%d:%d\n",i,sg[i]);    }        return 0;}

AC代码。

#include <iostream>#include <cstdio>#include <cstring>using namespace std;int main (){    int t;    scanf("%d",&t);    while (t--)    {        int n,q;        scanf("%d",&n);        int ans=0;        for (int i=1; i<=n; i++)        {            scanf("%d",&q);            if (q%8==0)                ans^=(q-1);            else if(q%8==7)                ans^=(q+1);            else                ans^=q;        }        if(ans==0)            printf ("Second player wins.\n");        else            printf ("First player wins.\n");    }    return 0;}