Codeforces 1037F. Maximum Reduction

#include<iostream>
#include<cstdio> 
#include<cmath>
#include<queue>
#include<vector>
#include<string.h>
#include<cstring>
#include<algorithm>
#include<set>
#include<map>
#include<fstream>
#include<cstdlib>
#include<ctime>
#include<list>
#include<climits>
#include<bitset>
using namespace std;
#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define fopen freopen("input.in", "r", stdin);freopen("output.in", "w", stdout);
#define left asfdasdasdfasdfsdfasfsdfasfdas1
#define tan asfdasdasdfasdfasfdfasfsdfasfdas
typedef long long ll;
typedef unsigned int un;
const int desll[4][2]={{0,1},{0,-1},{1,0},{-1,0}};
const ll mod=1e9+7;
const int maxn=2e6+7;
const int maxm=1e9+7;
const double eps=1e-4;
int m,n,k;
int ar[maxn];
int pre[maxn],nex[maxn];
int main()
{
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++)scanf("%d",&ar[i]);
    pre[1]=0;
    for(int i=2;i<=n;i++){
        int ins=i-1;
        while(ins>0 && ar[ins]<ar[i])ins=pre[ins];
        pre[i]=ins;
    }
    nex[n]=n+1;
    for(int i=n-1;i>0;i--){
        int ins=i+1;
        while(ins<=n && ar[ins]<=ar[i])ins=nex[ins];
        nex[i]=ins;
    }
    //for(int i=1;i<=n;i++)cout<<pre[i]<<" ";cout<<endl;
    //for(int i=1;i<=n;i++)cout<<nex[i]<<" ";cout<<endl;
    ll ans=0;
    for(int i=1;i<=n;i++){
        int l=pre[i],r=nex[i];
        ll a,b;
        if(i-l<r-i){
            for(int x=l+1;x<=i;x++){//x为符合条件的区间的起始点
                a=(i-x+1-1+m-2)/(m-1);//a为包含i的区间中最少的（m-1）段数
                a=max(a,1LL);
                b=(r-x-1)/(m-1);//b为以r-1为区间终点，可以贡献的最多(m-1)段数
                ///cout<<"x = "<<x<<" "<<a<<" "<<b<<endl;
                if(b>=a){
                    ans = (ans + (b-a+1)*ar[i]%mod)%mod;
                }
            }
        }
        else{
            for(int x=i;x<r;x++){
                a=(x-i+1-1+m-2)/(m-1);
                a=max(a,1LL);
                b=(x-l-1)/(m-1);
                //cout<<"x = "<<x<<" "<<a<<" "<<b<<endl;
                if(b>=a){
                    ans = (ans + (b-a+1)*ar[i]%mod)%mod;
                }
            }
        }
        //cout<<i<<" "<<ans<<" "<<a<<" "<<b<<endl;
    }
    printf("%I64d\n",ans);



    return 0;
}