P4301 [CQOI2013] 新Nim游戏

nim游戏的sg函数\(sg[x]=x\)，故，先手必败的条件是\(sg[a[1]]xorsg[a[2]]..xorsg[a[n]]==0\)故，我们需要第三次开始时，sg函数不为零。也即是找到一个集合，使得他的异或值不为零。我们采用线性基插入，无法插入时表示，异或值为零。又因为要找最小值，我们就需要将线性基所找到的集合拥有最大的权值，故排序后，从后往前插入即可。

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<cstdlib>
#include<climits>
#include<stack>
#include<vector>
#include<queue>
#include<set>
#include<bitset>
#include<map>
//#include<regex>
#include<cstdio>
#include <iomanip>
#pragma GCC optimize(2)
#define up(i,a,b)  for(int i=a;i<b;i++)
#define dw(i,a,b)  for(int i=a;i>b;i--)
#define upd(i,a,b) for(int i=a;i<=b;i++)
#define dwd(i,a,b) for(int i=a;i>=b;i--)
//#define local
typedef long long ll;
typedef unsigned long long ull;
const double esp = 1e-6;
const double pi = acos(-1.0);
const int INF = 0x3f3f3f3f;
const int inf = 1e9;
using namespace std;
ll read()
{
	char ch = getchar(); ll x = 0, f = 1;
	while (ch<'0' || ch>'9') { if (ch == '-')f = -1; ch = getchar(); }
	while (ch >= '0' && ch <= '9') { x = x * 10 + ch - '0'; ch = getchar(); }
	return x * f;
}
typedef pair<int, int> pir;
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
#define lrt root<<1
#define rrt root<<1|1
const int N = 1e3 + 10;
int n;
struct node {
	ll num, mag;
	bool operator<(const node a)
	{
		return num< a.num;
	}
}stn[N];
struct xxj {
	ll b[100];
	xxj()
	{
		upd(i, 0, 60)b[i] = 0;
	}
	bool insert(ll x)
	{
		dwd(i, 60, 0)
		{
			if (x >> i & 1)
			{
				if (b[i])
				{
					x ^= b[i];
				}
				else {
					b[i] = x;
					upd(j, 0, i - 1)if (b[i] >> j & 1)b[i] ^= b[j];
					upd(j, i + 1, 60)if (b[j] >> i & 1)b[j] ^= b[i];
					return 1;
				}
			}
		}
		return 0;
	}
}X;
int main()
{
	n = read();
	ll sum = 0;
	upd(i, 1, n)
	{
		stn[i].num = read(); 
		sum += stn[i].num;
	}
	sort(stn + 1, stn + 1 + n);
	ll ans = 0;
	dwd(i, n, 1)
	{
		if(X.insert(stn[i].num))
		{
			ans += stn[i].num;
		}
	}
	cout << sum-ans << endl;
	return 0;
}