题面
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.
题意
对于每一个k∈[1，n]，找出数列在任意连续k个数中都出现的数的最小值
分析
若某个数满足在连续k个数中出现，则任意两个这个数之间的距离小于等于k；
（当然也要考虑到这个数与起始和结尾的位置）
那么可以先处理每个数的相邻两数的最大距离（要特殊处理一下第一个和最后一个与边界的距离）
最后贪心的从小到大枚举即可
代码
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
inline int read() {
register int x = 0, f = 1;
register char c = getchar();
while (c < '0' || c > '9') {if (c == '-') f = -1; c = getchar();}
while (c >= '0' && c <= '9') x = (x << 3) + (x << 1) + c - '0',c = getchar();
return x * f;
}
const int MAXN = 3e5 + 1;
int n;
int a[MAXN], last[MAXN], dis[MAXN], ans[MAXN];
int main() {
int t = read();
while (t--){
memset(last, 0, sizeof(last));
memset(dis, 0, sizeof(dis));
memset(ans, -1, sizeof(ans));
n = read();
for (int i(1); i <= n; i++) a[i] = read();
for (int i(1); i <= n; i++){
dis[a[i]] = max(dis[a[i]], i - last[a[i]]);
last[a[i]] = i;
}
for (int i(1); i <= n; i++)
dis[i] = max(dis[i], n + 1 - last[i]);
for (int i(1); i <= n; i++)
for (int j(dis[i]); j <= n && ans[j] == -1; j++)//举个例子 对于1如果k = 3可以，那么k更大时也可以
ans[j] = i;
for (int i(1); i <= n; i++)
printf("%d ",ans[i]);
puts("");
}
return 0;
}