A Simple Nim

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

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

2
2
4 4
3
1 2 4

Sample Output

Second player wins.
First player wins.

Author

UESTC

Source

2016 Multi-University Training Contest 6

题意：跟普通的nim不一样的是，一堆可以分成三小堆；

思路：sg函数打表找规律；

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<set>
#include<map>
#include<queue>
#include<stack>
#include<vector>
using namespace std;
typedef long long ll;
#define PI acos(-1.0)
const int N=1e3+100,maxm=1e5+100,inf=0x3f3f3f3f,mod=1e9+7;
const ll INF=1e18+7;

int sg[N],mex[N];
void initsg()
{
    for(int i=1;i<=500;i++)
    {
        memset(mex,0,sizeof(mex));
        for(int j=1;j<=i-2;j++)
        {
            for(int k=1;k+j<i;k++)
            {
                int l=i-j-k;
                mex[sg[l]^sg[j]^sg[k]]=1;
            }
        }
        for(int j=1;j<=i;j++)
            mex[sg[j]]=1;
        for(int j=1;;j++)
        if(!mex[j])
        {
            sg[i]=j;
            break;
        }
    }
    for(int i=1;i<=500;i++)
    if(sg[i]!=i)cout<<sg[i]<<" "<<i<<endl;
}
int SG(int x)
{
    if(x%8==7)return x+1;
    if(x%8==0)return x-1;
    return x;
}
int main()
{
    //initsg();
    int T;
    scanf("%d",&T);
    while(T--)
    {
        int n;
        scanf("%d",&n);
        int ans=0;
        for(int i=1;i<=n;i++)
        {
            int x;
            scanf("%d",&x);
            ans^=SG(x);
        }
        if(ans)printf("First player wins.\n");
        else printf("Second player wins.\n");
    }
    return 0;
}