一道超级经典的单调栈问题，本题的关键在于你不要同时两边取等号，不然相等的区间会重复计算

还有记得开long long

#include<bits/stdc++.h>
using namespace std;
using ll=long long;
const int N = 2e5+10;
int n;
ll a[N];
ll l[N];
ll r[N];
int main()
{
	ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
	cin>>n;
	for(int i=1;i<=n;i++)cin>>a[i];
	a[0] = a[n+1] = -0x3f3f3f3f;
	
	
	stack<int>stk;
	stk.push(0);
	

	for(int i=1;i<=n;i++){
		while(stk.size()&&a[i]<=a[stk.top()])
			stk.pop();	
		l[i] = i-stk.top();
		stk.push(i);
		
	}
	
	stk = stack<int>();
	
	stk.push(n+1);
	for(int i=n;i>=1;i--){
		while(stk.size()&&a[i]<a[stk.top()])stk.pop();
		r[i] = stk.top()-i;
		stk.push(i);
	}
	
	
	
	
	ll res = 0;
	for(int i=1;i<=n;i++)
	  res = res + l[i]*r[i]*a[i];
	  
	cout<<res<<"\n";
	 
	
	return 0;
}