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 << "]" << "
";
    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, 1, BUF, stdin);
    return *bg++;
} 

inline int ri(){
    int x = 0, f = 1, c = gc();
    for(; c<48||c>57; f = c=='-'?-1:f, c=gc());
    for(; c>47&&c<58; x = x*10 + c - 48, 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-10;
const LL inf = 0x7fffffff;
const LL infLL = 0x7fffffffffffffffLL;
const LL mod = 1e9 + 7;
const int maxN = 3e5 + 7;
const LL ONE = 1;
const LL evenBits = 0xaaaaaaaaaaaaaaaa;
const LL oddBits = 0x5555555555555555;

LL N, M, K;

int main(){
    INIT(); 
    cin >> M >> K;
    N = (1 << M) - 1;

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