题目：
You are given an array a consisting of n integers numbered from 1 to n.
Let’s define the k-amazing number of the array as the minimum number that occurs in all of the subsegments of the array having length k (recall that a subsegment of a of length k is a contiguous part of a containing exactly k elements). If there is no integer occuring in all subsegments of length k for some value of k, then the k-amazing number is −1.
For each k from 1 to n calculate the k-amazing number of the array a.
Input
The first line contains one integer t (1≤t≤1000) — the number of test cases. Then t test cases follow.
The first line of each test case contains one integer n (1≤n≤3⋅105) — the number of elements in the array. The second line contains n integers a1,a2,…,an (1≤ai≤n) — the elements of the array.
It is guaranteed that the sum of n over all test cases does not exceed 3⋅105.
Output
For each test case print n integers, where the i-th integer is equal to the i-amazing number of the array.
Example
Input
3
5
1 2 3 4 5
5
4 4 4 4 2
6
1 3 1 5 3 1
Output
-1 -1 3 2 1
-1 4 4 4 2
-1 -1 1 1 1 1
题解：
#include<bits/stdc++.h>
using namespace std;
int t,n;
int a[300005],d[300005],ls[300005],ans[300005];
int main()
{
cin>>t;
while(t--)
{
scanf("%d",&n);
for(int i=1;i<=n;i++)
{
scanf("%d",&a[i]);
ls[i]=d[i]=0,ans[i]=-1;
}
for(int i=1;i<=n;i++)
{
int x=a[i];
d[x]=max(d[x],i-ls[x]);
ls[x]=i;
}
for(int i=1;i<=n;i++)
{
d[i]=max(d[i],n-ls[i]+1);
for(int j=d[i];j<=n&&ans[j]==-1;j++) ans[j]=i;
}
for(int i=1;i<=n;i++) printf("%d ",ans[i]);
printf("\n");
}
return 0;
}