AtCoder ABC 126F XOR Matching

题目链接：https://atcoder.jp/contests/abc126/tasks/abc126_f

题目大意

　　给定两个整数 M 和 K ，用小于 2^M 的的所有自然数，每个两个，用这些数排成一个长度为 2^M+1 的序列，使得序列满足以下条件：

1. 每个自然数只能用 2 次。
2. 设序列为 a，$\forall_{i < j}\ 满足a[i] == a[j]$有 a[i] xor a[i + 1] xor……xor a[j] == K。

　　问这个序列是否存在，存在则输出任意一个，不存在输出 -1。

分析

　　找规律。

　　貌似很难，其实巨简单。

　　首先，如果 K >=  2^M ，那就不用考虑了，绝对异或不出来。

　　其次，当 M == 1 时，K 为 0 则可行，K 为 1 则不可行。

　　最后是 M > 1 的情况，这里举个 M = 3，K = 4 的例子，某一绝对正确的摆法如下：0,1,2,3,5,6,7,4,7,6,5,3,2,1,0,4，加粗的是 K，说服一下自己，在 M > 1 的时候都能这么摆（实际上确实能这么摆）。

代码如下

 #include <bits/stdc++.h>
 using namespace std;

 #define INIT() ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
 #define Rep(i,n) for (int i = 0; i < (n); ++i)
 #define For(i,s,t) for (int i = (s); i <= (t); ++i)
 #define rFor(i,t,s) for (int i = (t); i >= (s); --i)
 #define ForLL(i, s, t) for (LL i = LL(s); i <= LL(t); ++i)
 #define rForLL(i, t, s) for (LL i = LL(t); i >= LL(s); --i)
 #define foreach(i,c) for (__typeof(c.begin()) i = c.begin(); i != c.end(); ++i)
 #define rforeach(i,c) for (__typeof(c.rbegin()) i = c.rbegin(); i != c.rend(); ++i)

 #define pr(x) cout << #x << " = " << x << "  "
 #define prln(x) cout << #x << " = " << x << endl

 #define LOWBIT(x) ((x)&(-x))

 #define ALL(x) x.begin(),x.end()
 #define INS(x) inserter(x,x.begin())

 #define ms0(a) memset(a,0,sizeof(a))
 #define msI(a) memset(a,inf,sizeof(a))
 #define msM(a) memset(a,-1,sizeof(a))

 #define MP make_pair
 #define PB push_back
 #define ft first
 #define sd second

 template<typename T1, typename T2>
 istream &operator>>(istream &in, pair<T1, T2> &p) {
     in >> p.first >> p.second;
     return in;
 }

 template<typename T>
 istream &operator>>(istream &in, vector<T> &v) {
     for (auto &x: v)
         in >> x;
     return in;
 }

 template<typename T1, typename T2>
 ostream &operator<<(ostream &out, const std::pair<T1, T2> &p) {
     out << "[" << p.first << ", " << p.second << "]" << "\n";
     return out;
 }

 inline int gc(){
     static const int BUF = 1e7;
     static char buf[BUF], *bg = buf + BUF, *ed = bg;

     if(bg == ed) fread(bg = buf, , BUF, stdin);
     return *bg++;
 } 

 inline int ri(){
     int x = , f = , c = gc();
     for(; c<||c>; f = c=='-'?-:f, c=gc());
     for(; c>&&c<; x = x* + c - , c=gc());
     return x*f;
 }

 typedef long long LL;
 typedef unsigned long long uLL;
 typedef pair< double, double > PDD;
 typedef pair< int, int > PII;
 typedef pair< string, int > PSI;
 typedef set< int > SI;
 typedef vector< int > VI;
 typedef map< int, int > MII;
 typedef pair< LL, LL > PLL;
 typedef vector< LL > VL;
 typedef vector< VL > VVL;
 const double EPS = 1e-;
 const LL inf = 0x7fffffff;
 const LL infLL = 0x7fffffffffffffffLL;
 const LL mod = 1e9 + ;
 const int maxN = 3e5 + ;
 const LL ONE = ;
 const LL evenBits = 0xaaaaaaaaaaaaaaaa;
 const LL oddBits = 0x5555555555555555;

 LL N, M, K;

 int main(){
     INIT();
     cin >> M >> K;
     N = ( << M) - ;

     if(K > N) cout << - << endl;
     else if(M == ) {
         if(K == ) cout << - << endl;
         else cout << "0 0 1 1" << endl;
     }
     else {
         For(i, , N) if(i != K) cout << i << " ";
         cout << K << " ";
         rFor(i, N, ) if(i != K) cout << i << " ";
         cout << K << " ";
         cout << endl;
     }
     return ;
 }