https://codeforces.com/contest/1313/problem/C2
题意：给出一组数，使这组数满足任意ai不存在j<i<k,a[j] > a[i] < a[k],求满足该条件的这组数和的最大值？
解法：单调递增栈，扩展出以每个数为最小值的左右区间，同时dp可以算出1-i左区间以该值为最小值的最大和，右区间同理。
#include<bits/stdc++.h>
typedef long long ll ;
#define int ll
#define mod 1000000007
#define gcd __gcd
#define rep(i , j , n) for(int i = j ; i <= n ; i++)
#define red(i , n , j) for(int i = n ; i >= j ; i--)
#define ME(x , y) memset(x , y , sizeof(x))
//ll lcm(ll a , ll b){return a*b/gcd(a,b);}
//ll quickpow(ll a , ll b){ll ans=1;while(b){if(b&1)ans=ans*a%mod;b>>=1,a=a*a%mod;}return ans;}
//int euler1(int x){int ans=x;for(int i=2;i*i<=x;i++)if(x%i==0){ans-=ans/i;while(x%i==0)x/=i;}if(x>1)ans-=ans/x;return ans;}
//const int N = 1e7+9; int vis[n],prime[n],phi[N];int euler2(int n){ME(vis,true);int len=1;rep(i,2,n){if(vis[i]){prime[len++]=i,phi[i]=i-1;}for(int j=1;j<len&&prime[j]*i<=n;j++){vis[i*prime[j]]=0;if(i%prime[j]==0){phi[i*prime[j]]=phi[i]*prime[j];break;}else{phi[i*prime[j]]=phi[i]*phi[prime[j]];}}}return len}
#define SC scanf
#define INF 0x3f3f3f3f
#define PI acos(-1)
#define pii pair<int,int>
#define fi first
#define se second
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
using namespace std;
const int maxn = 5e5+9;
const int N = 10000 ;
int s[maxn] , top ;
int a[maxn] , l[maxn] , r[maxn];
void solve(){
int n ;
cin >> n ;
rep(i , 1 , n){
cin >> a[i];
}
a[n+1] = a[0] = -INF;
rep(i , 1 , n+1){
while(top && a[s[top]] >= a[i]) top--;
if(!top) l[i] = a[i] * i ;
else l[i] = l[s[top]] + (i - s[top]) * a[i];
s[++top] = i ;
}
top = 0 ;
red(i , n , 0){
while(top && a[s[top]] >= a[i]) top--;
if(!top) r[i] = a[i] * (n-i+1) ;
else r[i] = r[s[top]] + (s[top] - i) * a[i];
s[++top] = i ;
}
int id = -1 , ans = -INF;
rep(i , 1 , n){
int temp = l[i] + r[i] - a[i];
if(temp > ans){
id = i ;
ans = temp;
}
}
red(i , id-1 , 1) a[i] = min(a[i+1] , a[i]);
rep(i , id+1 , n) a[i] = min(a[i] , a[i-1]);
rep(i , 1 , n) cout << a[i] << " " ;
}
signed main()
{
ios::sync_with_stdio(false);
cin.tie(0); cout.tie(0);
//int t ;
//cin >> t ;
//while(t--){
solve();
//}
}