AT2060 [AGC005B] Minimum Sum

题面

给你一个长为 \(n\) 的数列 \(a\)，求

\[\sum_{l=1}^{n}{\sum_{r=l}^{n}{min\{a_l \sim a_r\}}} \]

思路

考试题改的。

可以用单调栈算出贡献区间，然后乘法原理计算出总贡献。

具体见代码：

代码

// O(n)
#include <bits/stdc++.h>
#define int long long
using namespace std;

int a[300005];
int l[300005], r[300005];
int ll[300005], rr[300005];
int n;
int top, top2;
int st[300005];
int st2[300005];
long long ans;
//int ans1,ans2;
signed main() {
	//freopen("sequence.in", "r", stdin);
	//freopen("sequence.out", "w", stdout);
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	cin >> n;
	for (int i = 1; i <= n; i++) {
		cin >> a[i];
		r[i] = n + 1;
		rr[i] = n + 1;
	}
	for (int i = 1; i <= n; ++i) {
		while (top && a[st[top]] > a[i]) {
			r[st[top--]] = i;
		}
		l[i] = st[top];
		st[++top] = i;
	}
	for (int i = 1; i <= n; i++) {
		ans += (r[i] - i) * (i - l[i]) * a[i];
		//ans += (rr[i] - i) * (i - ll[i]) * a[i];
	}
	cout << ans << '\n';
	return 0;
}

标签：sequence,int,Sum,AT2060,long,st,Minimum,300005,top
来源： https://www.cnblogs.com/xiezheyuan/p/note-rem-at2060.html