Minimum Sum

题目描述

One day, Snuke was given a permutation of length N, a1,a2,…,aN, from his friend.

Find the following:

Constraints
1≤N≤200,000
(a1,a2,…,aN) is a permutation of (1,2,…,N).

输入

The input is given from Standard Input in the following format:
N
a1 a2 … aN

输出

Print the answer.
Note that the answer may not fit into a 32-bit integer.

样例输入

3
2 1 3

样例输出

9

这是一道单调栈 
单调栈的总结：https://blog.csdn.net/wubaizhe/article/details/70136174

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 2e5+;
struct node
{
    ll pre;
    ll net;
    ll num;
};
stack<node>sp;
ll s[maxn];
int main()
{
    ios::sync_with_stdio(false);
    int n;
    cin>>n;
    for(int i=;i<=n;i++) cin>>s[i];
    ll ans=;
    node fr,to;
    fr.net=fr.pre=;
    fr.num=s[];
    sp.push(fr);
    for(int i=;i<=n;i++){
        to.num=s[i];
        to.pre=to.net=;
        while(!sp.empty()&&to.num<=sp.top().num){
            fr=sp.top();
            sp.pop();
            if(!sp.empty()) sp.top().net+=fr.net;
            to.pre+=fr.pre;
            ans+=fr.pre*fr.num*fr.net;
        }
        sp.push(to);
    }
    while(!sp.empty())
    {
        fr=sp.top();
            sp.pop();
            if(!sp.empty()) sp.top().net+=fr.net;
            ans+=fr.pre*fr.num*fr.net;
    }
    cout<<ans<<endl;
    return ;
}

单调栈对求前后的有多少比他大/小很高效

1.首先这个运算要简化成：前面的大值*后面的大值（紧接的并且加上本身）

2.然后就是栈，极其烧脑┭┮﹏┭┮