Inversion

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题目链接

题意

计算

Code

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#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#ifndef ONLINE_JUDGE
#define dbg(x...) do{cout << "\033[33;1m" << #x << "->" ; err(x);} while (0)
void err(){cout << "\033[39;0m" << endl;}
template<template<typename...> class T, typename t, typename... A>
void err(T<t> a, A... x){for (auto v: a) cout << v << ' '; err(x...);}
template<typename T, typename... A>
void err(T a, A... x){cout << a << ' '; err(x...);}
#else
#define dbg(...)
#endif
#define inf 1ll << 50
const int N = 1e5 + 5;

int lcy[N][20];
int a[N];

void rmq_pre(int n)
{
	for (int i = 1; i <= n; i++)
		lcy[i][0] = a[i];
	for (int j = 1; (1 << j) <= n; j++)
		for (int i = 1; i + (1 << j) <= n + 1; i++)
		{
			lcy[i][j] = max(lcy[i][j - 1], lcy[i + (1 << (j - 1))][j - 1]);
		}
}

int rmq(int l, int r)
{
	int k = 31 - __builtin_clz(r - l + 1);
//	dbg(l, k, r - (1 << k) + 1, 1 << k);
	return max(lcy[l][k], lcy[r - (1 << k) + 1][k]);
}

int main()
{
	int n, T;
	scanf("%d", &T);
	while (T--)
	{
		scanf("%d", &n);
		for (int i = 1; i <= n; i++)
			scanf("%d", &a[i]);
		rmq_pre(n);
		
		for (int i = 2; i <= n; i++)
		{
			int tmp = 0;
			for (int l = 0, r = i; l < n; l = r, r += i)
			{
				//dbg(l, r);
				tmp = max(tmp, rmq(l + 1, min(r - 1, n)));
			}
			printf("%d%c", tmp, i == n? '\n' : ' ');
		}
		
//		dbg(rmq(4, 5), lcy[4][1]);
	}
	return 0;
}
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