题目
思路
先4个单调栈求出每个位置能够贡献的区间然后一波统计直接AC。
code:
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<string>
#include<cstring>
#include<cmath>
#include<queue>
#include<deque>
#define int long long
using namespace std;
int l[1000101],r[1000101],l2[1000101],r2[1000101],a[1000101],n,ans;
int y[1000101],t=1;
signed main()
{
cin>>n;
y[t]=0;
a[n+1]=a[0]=0x7ffffff;
for (int i=1;i<=n;i++)
{
cin>>a[i];
while (a[i]>=a[y[t]]) t--;
l[i]=y[t];
y[++t]=i;
}
t=1,y[t]=n+1;
for (int i=n;i>=1;i--)
{
while (t>0&&a[i]>a[y[t]]) t--;
r[i]=y[t];
y[++t]=i;
}
a[n+1]=a[0]=-0x7ffffff;
t=1,y[t]=n+1;
for (int i=n;i>=1;i--)
{
while (a[i]<=a[y[t]]) t--;
r2[i]=y[t];
y[++t]=i;
}
t=1,y[t]=0;
for (int i=1;i<=n;i++)
{
while (t>0&&a[i]<a[y[t]]) t--;
l2[i]=y[t];
y[++t]=i;
ans=ans+a[i]*(i-l[i])*(r[i]-i);
ans=ans-a[i]*(i-l2[i])*(r2[i]-i);
}
cout<<ans;
return 0;
}