abc281E Least Elements

给定数组a[n]以及两个数m与k，求数组a[n]上大小为m的滑动窗口里最小的k个数之和。
1<=k<=m<=n<=2E5; 1<=a[i]<=1E9

分析：直接套带前缀和的平衡树即可。

#include <bits/stdc++.h>
using i64 = long long;

// Treap模板...

const int N = 200005;
int n, m, k, a[N];

void solve() {
    std::cin >> n >> m >> k;
    for (int i = 1; i <= n; i++) {
        cin >> a[i];
    }
    Treap<int> tp;
    for (int i = 1; i <= m - 1; i++) {
		tp.insert(a[i]);
	}
	for (int i = m; i <= n; i++) {
        tp.insert(a[i]);
        std::cout << tp.ksum(k) << " ";
        tp.erase(a[i - m + 1]);
    }
}

int main() {
    std::cin.tie(0)->sync_with_stdio(0);
    int t = 1;
    while (t--) solve();
    return 0;
}