HDU 5795 A Simple Nim (SG函数+打表找规律)

A Simple Nim

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1129    Accepted Submission(s): 637

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

POINT：

游戏的和的SG函数值是它的所有子游戏的SG函数值的异或（重要！！）

#include <iostream>#include <cstdio>#include <cstring>using namespace std;#define ll long long/*int sg[10005]= {0,1,2};//打表int vis[10005];void play(){    for(int i=3;i<=100;i++)    {        memset(vis,0,sizeof vis);        for(int j=1;j<=i;j++)        {            for(int k=1;k<=i;k++)            {                if(j+k<i)                {                    vis[sg[j]^sg[k]^sg[i-j-k]]=1;                }            }        }        for(int j=0;j<i;j++)        {            vis[sg[j]]=1;        }        for(int j=0;;j++)        {            if(!vis[j])            {                sg[i]=j;                break;            }        }    }    for(int i=0;i<=100;i++)    {        printf("%d %d\n",i,sg[i]);    }}*/ll sg(ll n){    if(n%8==0) return n-1;    else if((n+1)%8==0) return n+1;    else return n;}int main(){   // play();    int T;    scanf("%d",&T);    while(T--)    {        int n;        scanf("%d",&n);        ll res=0;        while(n--)        {            ll a;            scanf("%lld",&a);            res=res^sg(a);        }        if(res==0) printf("Second player wins.\n");        else printf("First player wins.\n");            }    return 0;}