Atcoder ABC140-E- Second Sum

Second Sum

Input



Input is given from Standard Input in the following format:

Output

Sample Input


Sample Output

Note

#include<bits/stdc++.h>
#define me(a,x) memset(a,x,sizeof(a))
#define IN freopen("in.txt","r",stdin);
#define OUT freopen("out.txt","w",stdout);
#define sc scanf
#define itn int
#define STR clock_t startTime = clock();
#define END clock_t endTime = clock();cout << double(endTime - startTime) / CLOCKS_PER_SEC *1000<< "ms" << endl;
using namespace std;
const int N=2e5+5;
const long long mod=1e9+7;
const long long mod2=998244353;
const int oo=0x7fffffff;
const int sup=0x80000000;
typedef long long ll;
typedef unsigned long long ull;
template <typename it>void db(it *begin,it *end){while(begin!=end)cout<<(*begin++)<<" ";puts("");}
template <typename it>
string to_str(it n){string s="";while(n)s+=n%10+'0',n/=10;reverse(s.begin(),s.end());return s;}
template <typename it>int o(it a){cout<<a<<endl;return 0;}
inline ll mul_64(ll x,ll y,ll c){return (x*y-(ll)((long double)x/c*y)*c+c)%c;}
inline ll ksm(ll a,ll b,ll c){ll ans=1;for(;b;b>>=1,a=a*a%c)if(b&1)ans=ans*a%c;return ans;}
inline void exgcd(ll a,ll b,ll &x,ll &y){if(!b)x=1,y=0;else exgcd(b,a%b,y,x),y-=x*(a/b);}
int a[N],n;
int lpos[N],rpos[N];
int pos[N];
int main(){
   // IN
    cin>>n;
    for(int i=1;i<=n;i++)sc("%d",&a[i]),pos[a[i]]=i;
    deque<int>q,p;
    q.push_back(a[1]);lpos[a[1]]=0;
    for(int i=2;i<=n;i++){
        while(!q.empty()&&q.back()<a[i])q.pop_back();
        if(q.empty())lpos[a[i]]=0;
        else {
            int x=q.back();
            lpos[a[i]]=pos[x];
        }
        q.push_back(a[i]);
    }
    p.push_back(a[n]);rpos[a[n]]=n+1;
    for(int i=n-1;i>=1;i--){
        while(!p.empty()&&p.back()<a[i])p.pop_back();
        if(p.empty())rpos[a[i]]=n+1;
        else {
            int x=p.back();
            rpos[a[i]]=pos[x];
        }
        p.push_back(a[i]);
    }
    ll ans=0;
    for(int i=1;i<n;i++){
        int l=lpos[i],r=rpos[i];
        if(i==1){
            if(l==0)ans+=1;
            else if(r==n+1)ans+=1;
            else ans+=2;
            continue;
        }
        int ql=0,qr=n+1;
        for(int j=l-1;j>=1;j--)if(a[j]>i){ql=j;break;}
        for(int j=r+1;j<=n;j++)if(a[j]>i){qr=j;break;}
        if(l==0){
            ans+=1LL*i*(pos[i]-l)*(qr-r);
        }else if(r==n+1){
            ans+=1LL*i*(r-pos[i])*(l-ql);
        }else{
            ans+=1LL*i*(r-pos[i])*(l-ql);
            ans+=1LL*i*(pos[i]-l)*(qr-r);
        }
    }
    o(ans);
}