[CF1037F]sequence

The meaning of problems

https://codeforces.com/contest/1037/problem/F

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Think

Excerpt from a more interesting approach. We divide and conquer sequence, each statistical contribution across the mid section of. Which is to ensure the accuracy: only each interval corresponds to a mid. For the left interval, calculate its maximum extension.

Next, from left to right and right to enumerate every point range. For point $ i $, the right end of the span section above it is a $ k-1 $. Therefore, in addition to the right of the maximum value of the same left block intervals, the remaining prefix contribution to span (maximum extension) to (k-1) determined by the left and interval.

Code is small, but a lot of details. Although excellent algorithm complexity is $ O (nlogn) $, but after all, to solve this kind of problem.

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Code

 1 #include<bits/stdc++.h>
 2 #define mod 1000000007
 3 using namespace std;
 4 typedef long long int ll;
 5 const int maxn=1E6+5;
 6 int n,k;
 7 ll ans,a[maxn],maxx[maxn],sum[maxn];
 8 inline ll max(ll x,ll y)
 9 {
10     return x>y?x:y;
11 }
12 void solve(int l,int r)
13 {
14     if(r-l+1<k)
15         return;
16     int mid=(l+r)>>1;
17     maxx[mid]=a[mid];
18     for(int i=mid-1;i>=l;--i)
19         maxx[i]=max(maxx[i+1],a[i]);
20     sum[l-1]=0;
21     sum[l]=maxx[l];
22     for(int i=l+1;i<=mid;++i)
23     {
24         sum[i]=0;
25         int pos=i-k+1;
26         if(pos>=l)
27             sum[i]=sum[pos];
28         sum[i]=(sum[i]+maxx[i])%mod;
29     }
30     ll now=a[mid+1];
31     for(int i=mid+1,j=mid;i<=r;++i,now=max(now,a[i]))
32     {
33         if(i-l+1<k)
34             continue;
35         while(j>=l&&maxx[j]<now)
36             --j;
37         int len=i-j+1;
38         int block=(len-2)/(k-1);
39         int out=block+1;
40         block-=(i-mid-1)/(k-1);
41         ans=(ans+block*now%mod+sum[max(l-1,i-out*(k-1))])%mod;
42     }
43     solve(l,mid),solve(mid+1,r);
44 }
45 int main()
46 {
47     ios::sync_with_stdio(false);
48     cin>>n>>k;
49     for(int i=1;i<=n;++i)
50         cin>>a[i];
51     solve(1,n);
52     cout<<ans<<endl;
53     return 0;
54 }

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