【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=,inv2=(MOD+)/;
       double eps=1e-;
       int INF=1e9;

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

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

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