Alice has a magic array. She suggests that the value of a interval is equal to the sum of the values in the interval, multiplied by the smallest value in the interval.
Now she is planning to find the max value of the intervals in her array. Can you help her?
Input
First line contains an integer n(1 le n le 5 imes 10 ^5n(1≤n≤5×105).
Second line contains nn integers represent the array a (-10^5 le a_i le 10^5)a(−105≤ai≤105).
Output
One line contains an integer represent the answer of the array.
样例输入
5 1 2 3 4 5
样例输出
36
用单调栈判断以每个值为最小值的最大左边界和右边界。后对对每个值分成负数和正数讨 论取可行区间内的最小或最大值，方法为求前缀和和后缀和，然后用线段树求区间最值。
线段树维护的时候，左区间维护后缀和，右区间维护前缀和，找的时候，在i值左边找后缀，在i值右边找前缀，然后交叉的部分就是满足的区间。
代码:
1 //I-线段树+单调栈 2 #include<bits/stdc++.h> 3 using namespace std; 4 typedef long long ll; 5 const int maxn=5e5+10; 6 const int inf=0x3f3f3f3f; 7 8 #define lson l,m,rt<<1 9 #define rson m+1,r,rt<<1|1 10 11 int a[maxn],l[maxn],r[maxn]; 12 ll pre[maxn],beh[maxn],maxl[maxn<<2],minl[maxn<<2],maxr[maxn<<2],minr[maxn<<2]; 13 14 void pushup(int rt) 15 { 16 maxl[rt]=max(maxl[rt<<1],maxl[rt<<1|1]); 17 minl[rt]=min(minl[rt<<1],minl[rt<<1|1]); 18 maxr[rt]=max(maxr[rt<<1],maxr[rt<<1|1]); 19 minr[rt]=min(minr[rt<<1],minr[rt<<1|1]); 20 } 21 22 void build(int l,int r,int rt) 23 { 24 if(l==r){ 25 maxl[rt]=minl[rt]=beh[l]; 26 maxr[rt]=minr[rt]=pre[l]; 27 return ; 28 } 29 30 int m=(l+r)>>1; 31 build(lson); 32 build(rson); 33 pushup(rt); 34 } 35 36 ll query(int op,int L,int R,int l,int r,int rt) 37 { 38 if(L<=l&&r<=R){ 39 if (op==1) return maxl[rt]; 40 else if(op==2) return minl[rt]; 41 else if(op==3) return maxr[rt]; 42 else if(op==4) return minr[rt]; 43 } 44 45 int m=(l+r)>>1; 46 ll ret; 47 if(op==1||op==3){ 48 ret=-inf; 49 if(L<=m) ret=max(ret,query(op,L,R,lson)); 50 if(R> m) ret=max(ret,query(op,L,R,rson)); 51 } 52 else if(op==2||op==4){ 53 ret=inf; 54 if(L<=m) ret=min(ret,query(op,L,R,lson)); 55 if(R> m) ret=min(ret,query(op,L,R,rson)); 56 } 57 return ret; 58 } 59 60 deque<int> deq;//因为是双端队列，所以插的时候要插到头上才能实现栈的功能 61 62 int main() 63 { 64 int n; 65 scanf("%d",&n); 66 for(int i=1;i<=n;i++){ 67 scanf("%d",&a[i]); 68 } 69 for(int i=1;i<=n;i++){ 70 pre[i]=pre[i-1]+a[i]; 71 } 72 for(int i=n;i>=1;i--){ 73 beh[i]=beh[i+1]+a[i]; 74 } 75 build(1,n,1); 76 for(int i=1;i<=n;i++){ 77 while(deq.size()&&a[deq.front()]>=a[i]) deq.pop_front(); 78 if(deq.empty()) l[i]=1; 79 else l[i]=deq.front()+1; 80 deq.push_front(i); 81 } 82 deq.clear(); 83 for(int i=n;i>=1;i--){ 84 while(deq.size()&&a[deq.front()]>=a[i]) deq.pop_front(); 85 if(deq.empty()) r[i]=n; 86 else r[i]=deq.front()-1; 87 deq.push_front(i); 88 } 89 ll maxx=-inf,ret; 90 for(int i=1;i<=n;i++){ 91 if(a[i]>=0){ 92 ret=query(1,l[i],i,1,n,1); 93 ret+=query(3,i,r[i],1,n,1); 94 ret=ret-beh[i]-pre[i]+a[i]; 95 ret*=a[i]; 96 // cout<<query(1,l[i],i,1,n,1)<<" "<<query(3,i,r[i],1,n,1)<<endl; 97 } 98 else{ 99 ret=query(2,l[i],i,1,n,1); 100 ret+=query(4,i,r[i],1,n,1); 101 ret=ret-beh[i]-pre[i]+a[i]; 102 ret*=a[i]; 103 } 104 maxx=max(maxx,ret); 105 } 106 printf("%lld ",maxx); 107 }