【HDOJ6614】AND Minimum Spanning Tree（签到）

题意：给定标号从1到n的n个点，链接两个点x，y的代价为x AND y，求最小生成树总代价与满足代价最小的前提下字典序最小的方案

n<=2e5

思路：

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned int uint;
typedef unsigned long long ull;
typedef pair<int,int> PII;
typedef pair<ll,ll> Pll;
typedef vector<int> VI;
#define N  4100
#define M  4100000
#define fi first
#define se second
#define MP make_pair
#define pi acos(-1)
#define mem(a,b) memset(a,b,sizeof(a))
#define rep(i,a,b) for(int i=(int)a;i<=(int)b;i++)
#define per(i,a,b) for(int i=(int)a;i>=(int)b;i--)
#define lowbit(x) x&(-x)
#define Rand (rand()*(1<<16)+rand())
#define id(x) ((x)<=B?(x):m-n/(x)+1)
#define ls p<<1
#define rs p<<1|1

const ll MOD=998244353,inv2=(MOD+1)/2;
      double eps=1e-6;
      int INF=1e9;


int read()
{
   int v=0,f=1;
   char c=getchar();
   while(c<48||57<c) {if(c=='-') f=-1; c=getchar();}
   while(48<=c&&c<=57) v=(v<<3)+v+v+c-48,c=getchar();
   return v*f;
}

int check(int x)
{
    int t=x&(-x);
    if(t==x) return 1;
    return 0;
}


int main()
{
    //freopen("1.in","r",stdin);
    //freopen("1.out","w",stdout);
    int cas=read();
    int mx=(1<<20)-1;
    while(cas--)
    {
        int n=read();
        if(check(n+1))
        {
            printf("1
");
            rep(i,2,n-1)
            {
                int t=i^mx;
                //printf("t=%d
",t);
                t=lowbit(t);
                //printf("lowbit=%d
",t);
                printf("%d ",t);
            }
            printf("1
");
        }
         else
         {
             printf("0
");
             rep(i,2,n)
             {
                int t=i^mx;
                t=lowbit(t);
                if(i<n) printf("%d ",t);
                 else printf("%d",t);
             }
             printf("
");
         }
    }
    return 0;
}