题意

给出一个大小为 $n$ 的数组 $a$，计算当 $k$ 从 $1$ 到 $n$ 取值时在所有 $k$ 长区间中出现的数的最小值。

题解

记录一个值两两间的最大距离，该距离的 $k$ 长区间及之后更长的区间都可能以该值为最小值。

Tips

注意两端的处理。

代码一

#include <bits/stdc++.h>
using namespace std;
int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int t;
    cin >> t;
    while (t--) {
        int n;
        cin >> n;
        vector<int> pre(n + 1), dis(n + 1);
        for (int i = 1; i <= n; i++) {
            int x;
            cin >> x;
            dis[x] = max(dis[x], i - pre[x]);
            pre[x] = i;
        }
        for (int i = 1; i <= n; i++) {
            if (pre[i]) {
                dis[i] = max(dis[i], n + 1 - pre[i]);
            }
        }
        vector<int> ans(n + 1, 1e9);
        for (int i = 1; i <= n; i++) {
            ans[dis[i]] = min(ans[dis[i]], i);
        }
        for (int i = 2; i <= n; i++) {
            ans[i] = min(ans[i], ans[i - 1]);
        }
        for (int i = 1; i <= n; i++) {
            cout << (ans[i] == 1e9 ? -1 : ans[i]) << " 
"[i == n];
        }
    }
    return 0;
}

代码二

#include <bits/stdc++.h>
using namespace std;
int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int t;
    cin >> t;
    while (t--) {
        int n;
        cin >> n;
        vector<vector<int>> pos(n + 1, {0});
        for (int i = 1; i <= n; i++) {
            int x;
            cin >> x;
            pos[x].push_back(i);
        }
        for (int i = 1; i <= n; i++) {
            pos[i].push_back(n + 1);
        }
        vector<int> dis(n + 1);
        for (int i = 1; i <= n; i++) {
            if (pos[i].size() > 2) {
                for (int j = 1; j < int(pos[i].size()); j++) {
                    dis[i] = max(dis[i], pos[i][j] - pos[i][j - 1]);
                }
            }
        }
        vector<int> ans(n + 1, 1e9);
        for (int i = 1; i <= n; i++) {
            ans[dis[i]] = min(ans[dis[i]], i);
        }
        for (int i = 2; i <= n; i++) {
            ans[i] = min(ans[i], ans[i - 1]);
        }
        for (int i = 1; i <= n; i++) {
            cout << (ans[i] == 1e9 ? -1 : ans[i]) << " 
"[i == n];
        }
    }
    return 0;
}

参考博客

https://www.cnblogs.com/petjelinux/p/13742723.html