http://cplusoj.com/d/senior/p/SS231102C

首先可以通过枚举逆序对点的贡献推出无解情况为 $n\mathrm{mod}4>1$

然后构造可以按 $n\mathrm{mod}3$ 进行分类

#include<bits/stdc++.h>
using namespace std;
#ifdef LOCAL#define debug(...) fprintf(stdout, ##__VA_ARGS__)
#else#define debug(...) void(0)
#endif
#define int long long
inline int read(){int x=0,f=1;char ch=getchar(); while(ch<'0'||
ch>'9'){if(ch=='-')f=-1;ch=getchar();}while(ch>='0'&&ch<='9'){
x=(x<<1)+(x<<3)+(ch^48);ch=getchar();}return x*f;}
#define Z(x) (x)*(x)
#define pb push_back
//mt19937 rand(time(0));
//mt19937_64 rand(time(0));
//srand(time(0));
#define N 2000010
//#define M
//#define mo
int n, m, i, j, k, T;
int a[N], b[N], mp[N]; 
vector<int>v[N]; void lian(int *a, int *b, int x, int y) {a[x]=y; b[y]=x; 
}void fenzhi(int n, int l, int r, int op, int *a, int *b) {
//	debug("%lld [%lld %lld] %lld\n", n, l, r, op); int x, i, j, k, p=l-1; if(!n) return ; if(n==1) return a[l]=b[l]=l, void(); if(op==0) {x=(n-1)/3; 		if(n<=8) {
//			debug(">> %lld\n", n); if(n!=4 && n!=5 && n!=8) return ; for(i=1; i<=n; ++i) lian(a, b, p+i, p+v[n][i]); 
//			for(i=l; i<=r; ++i) debug("%d ", a[i]); debug("\n"); 
//			for(i=l; i<=r; ++i) debug("%d ", b[i]); debug("\n"); return ; }if(n%3==1) {for(i=1, j=n; i<=x; ++i, --j) lian(a, b, i+p, j+p); for(i=2, j=n; i<=x+1; ++i, --j) lian(b, a, i+p, j+p); lian(a, b, x+1+p, 1+p); fenzhi(n-2*x-1, l+x+1, r-x, 0, a, b); return ; }if(n%3==2) {for(i=1, j=n; i<=x; ++i, --j) lian(a, b, i+p, j+p); for(i=2, j=n; i<=x+1; ++i, --j) lian(b, a, i+p, j+p); lian(b, a, 1+p, x+2+p); lian(a, b, x+3+p, r-x); lian(a, b, r-x, x+2+p); fenzhi(n-2*x-3, l+x+2, r-x-1, 1, a, b); 
//			debug("GG "); for(i=l; i<=r; ++i) debug("%d ", a[i]); debug("\n"); return ; }if(n%3==0) {for(i=1, j=n; i<=x; ++i, --j) lian(a, b, i+p, j+p); for(i=2, j=n; i<=x+1; ++i, --j) lian(b, a, i+p, j+p); 
//			debug("# "); for(i=l+x+1; i<=r-x; ++i) debug("%d ", a[i]); debug("\n"); fenzhi(n-2*x-1, l+x+1, r-x, 0, a, b); 
//			debug("# "); for(i=l+x+1; i<=r-x; ++i) debug("%d ", a[i]); debug("\n"); 
//			for(i=l+x+1; i<=r-x; ++i) debug("%d ", abs(a[i]-i)); debug("\n"); lian(a, b, x+1+p, r-x); lian(a, b, x+2+p, l); 
//			for(i=l; i<=r; ++i) debug("%d ", abs(a[i]-i)); debug("\n"); return ; }}if(op==1) {x=n/3-1; for(i=1, j=n; i<=x; ++i, --j) lian(b, a, i+p, j+p); for(i=2, j=n; i<=x+1; ++i, --j) lian(a, b, i+p, j+p); lian(a, b, l-2, x+1+p); if(n<=5) {if(n==4) {lian(a, b, l-2, l+1); lian(a, b, l+1, l+2); lian(a, b, l+2, l); lian(a, b, r, r); }if(n==5) {lian(a, b, l-2, l+1); lian(a, b, l+1, l+2); lian(a, b, l+2, r); lian(a, b, l+3, l+3); lian(a, b, r, l); }return ; }if(n%3==0) {fenzhi(n-2*x-1, l+x+1, r-x, 0, a, b); return ; }if(n%3==1) {lian(a, b, l-2, x+2+p); lian(a, b, x+2+p, r-x); lian(a, b, r-x, x+3+p); fenzhi(n-2*x-3, l+x+2, r-x-1, 1, b, a); return ; }if(n%3==2) {fenzhi(n-2*x-1, l+x+1, r-x, 0, b, a); lian(a, b, l-2, x+2+p); lian(a, b, r-x, x+1+p); return ; }}
}signed main()
{freopen("destruction.in", "r", stdin);freopen("destruction.out", "w", stdout);#ifdef LOCALfreopen("in.txt", "r", stdin);freopen("out.txt", "w", stdout);#endifv[4]={0, 4, 2, 1, 3}; v[5]={0, 5, 3, 1, 4, 2}; v[8]={0, 8, 7, 5, 4, 1, 3, 6, 2}; T=read();while(T--) {n=m=read(); if(n%4 && (n-1)%4) { printf("-1\n"); continue; }for(i=1; i<=n; ++i) a[i]=b[i]=i; fenzhi(n, 1, n, 0, a, b); 
//		memset(mp, 0, sizeof(mp));for(i=1; i<=n; ++i) printf("%d ", a[i]);
//		debug("\n"); 
//		for(i=1; i<=n; ++i) debug("%2d ", i);
//		debug("\n") ; 
//		for(i=1; i<=n; ++i) debug("%d ", abs(a[i]-i)); 
//		for(i=1; i<=n; ++i) if(mp[abs(a[i]-i)]) break; else mp[abs(a[i]-i)]=1; 
//		if(i<=n) printf("\nsdfasfsad asdf safasdf %lld\n", n); printf("\n"); }return 0;
}