【[AGC005B] Minimum Sum】 题解

题目链接

从1开始从小到大考虑，用set维护每个数左右的扩散范围，然后答案为这个数的区间就是左端点个数 $\times$ 右端点个数。

Code

// Problem: AT2060 [AGC005B] Minimum Sum
// Contest: Luogu
// URL: https://www.luogu.com.cn/problem/AT2060
// Memory Limit: 250 MB
// Time Limit: 2000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

#include<bits/stdc++.h>
using namespace std;
#define int long long
inline int read(){int x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;
ch=getchar();}while(ch>='0'&&ch<='9'){x=(x<<1)+
(x<<3)+(ch^48);ch=getchar();}return x*f;}
//#define mo
//#define M
//#define M
struct node
{
	int x, id; 
}a[200010]; 
int n, m, i, j, k; 
set<int>s; 
set<int>::iterator it;
set<int>::iterator iw; 
int ans, l, r; 

bool cmp(node x, node y)
{
	return x.x<y.x; 
}

signed main()
{
//	freopen("tiaoshi.in","r",stdin);
//	freopen("tiaoshi.out","w",stdout);
	n=read(); 
	for(i=1; i<=n; ++i) 
		a[i].x=read(), a[i].id=i; 
	sort(a+1, a+n+1, cmp); 
	s.insert(0); s.insert(n+1); 
	for(i=1; i<=n; ++i)
	{
		// printf("> %lld %lld\n", a[i].x, a[i].id); 
		it = s.upper_bound(a[i].id);
		r=*it;
		it--;
		l=*it; 
		k=max((long long)0, r-l-1); 
		++l; --r; 
		// printf("%lld %lld %lld\n", l, r, k); 
		ans+=a[i].x*max((long long)0, (a[i].id-l+1)*(r-a[i].id+1)); 
		s.insert(a[i].id); 
	}
	printf("%lld", ans); 
	return 0;
}