CodeForces - 287C Lucky Permutation(构造)

Posted 2021-09-14 Frozen_Guardian

题目链接：点击查看

题目大意：构造一个合法的排列，满足 $p_{p_i}=n-i+1$

题目分析：因为第四个样例的存在降低了本题的难度，不然感觉还是有点难度的一道题。

5
2 5 3 1 4

观察上面的这个构造方法，不难看出可以令四个数组成一个环，这样当 $n$ 只有是 $4$ 的倍数或者余数为 $1$ 时才有解，多出来的那个数放到中间就好啦

一组合法的环：$i\to i+1\to n-i+1\to n-i\to i$

代码：

// Problem: Lucky Permutation
// Contest: Virtual Judge - CodeForces
// URL: https://vjudge.net/problem/CodeForces-287C
// Memory Limit: 262 MB
// Time Limit: 2000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

// #pragma GCC optimize(2)
// #pragma GCC optimize("Ofast","inline","-ffast-math")
// #pragma GCC target("avx,sse2,sse3,sse4,mmx")
#include<iostream>
#include<cstdio>
#include<string>
#include<ctime>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<stack>
#include<climits>
#include<queue>
#include<map>
#include<set>
#include<sstream>
#include<cassert>
#include<bitset>
#include<list>
#include<unordered_map>
#define lowbit(x) (x&-x)
using namespace std;
typedef long long LL;
typedef unsigned long long ull;
template<typename T>
inline void read(T &x)
{
	T f=1;x=0;
	char ch=getchar();
	while(0==isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
	while(0!=isdigit(ch)) x=(x<<1)+(x<<3)+ch-'0',ch=getchar();
	x*=f;
}
template<typename T>
inline void write(T x)
{
	if(x<0){x=~(x-1);putchar('-');}
    if(x>9)write(x/10);
    putchar(x%10+'0');
}
const int inf=0x3f3f3f3f;
const int N=1e6+100;
int ans[N];
int main()
{
#ifndef ONLINE_JUDGE
//	freopen("data.in.txt","r",stdin);
//	freopen("data.out.txt","w",stdout);
#endif
//	ios::sync_with_stdio(false);
	int n;
	read(n);
	if(n%4!=0&&n%4!=1) {
		puts("-1");
		return 0;
	}
	if(n%4==1) {
		ans[n/2+1]=n/2+1;
	}
	for(int i=1;i<=n/2;i+=2) {
		ans[i]=i+1;
		ans[i+1]=n-i+1;
		ans[n-i+1]=n-i;
		ans[n-i]=i;
	}
	for(int i=1;i<=n;i++) {
		printf("%d ",ans[i]);
	}
	return 0;
}