[More than 2019 school] Make Rounddog Happy (divide and conquer)

Rounddog always has an array a1,a2,⋯,ana1,a2,⋯,an in his right pocket, satisfying 1≤ai≤n1≤ai≤n .

A subarray is a non-empty subsegment of the original array. Rounddog defines a good subarray as a subsegment al,al+1,⋯,aral,al+1,⋯,ar that all elements in it are different and max(al,al+1,…,ar)−(r−l+1)≤kmax(al,al+1,…,ar)−(r−l+1)≤k .

Rounddog is not happy today. As his best friend, you want to find all good subarrays of aa to make him happy. In this case, please calculate the total number of good subarrays of aa .

InputThe input contains several test cases, and the first line contains a single integer T (1≤T≤20)T (1≤T≤20) , the number of test cases.

The first line of each test case contains two integers n (1≤n≤300000)n (1≤n≤300000) and k (1≤k≤300000)k (1≤k≤300000) .

The second line contains nn integers, the ii -th of which is ai (1≤ai≤n)ai (1≤ai≤n) .

It is guaranteed that the sum of nn over all test cases never exceeds 10000001000000 .
OutputOne integer for each test case, representing the number of subarrays Rounddog likes.
Sample Input

2
5 3
2 3 2 2 5
10 4
1 5 4 3 6 2 10 8 4 5

Sample Output

7
31 
SOLUTION: 
it is easy to think of every number considered as the maximum value, then the process that he can think on both sides of the extended range 
, but well out to both sides, but it must be between the cross, then the autistic 
later learned that the partition can be good to engage in this problem 
solution to a problem: https: //www.cnblogs.com/Dillonh/p/11390927.html 
all things are heuristic complexity of metaphysics. . . . . qwq

#include<bits/stdc++.h>
#define ll long long
#define rep(i,a,b) for(int i=a;i<=b;i++)
using namespace std;
const int maxn=1000010;
int a[maxn],st[maxn][21],lg[maxn],K,N; ll ans;
int A[maxn],B[maxn],vis[maxn];
int get(int L,int R)
{
    int k=lg[R-L+1];
    return a[st[L][k]]>=a[st[R-(1<<k)+1][k]]?st[L][k]:st[R-(1<<k)+1][k];
}
void solve(int L,int R)
{
    if(L>R) return ;
    int pos=get(L,R);
    if(pos-L<R-pos){
        rep(i,L,pos){
            int t=a[pos]-K,lR=i+t-1;
            int fcy=min(R,B[i]);
            lR=max(lR,pos);
            if(lR>fcy) continue;
            ans+=fcy-lR+1;
        }
    }
    else {
        rep(i,pos,R){
            int t=a[pos]-K,rL=i-t+1;
            int fcy=max(L,A[i]);
            rL=min(rL,pos);
            if(rL<fcy) continue;
            ans+=rL-fcy+1;
        }
    }
    solve(L,pos-1); solve(pos+1,R);
}
int main()
{
    int T;
    lg[0]=-1; rep(i,1,maxn-1) lg[i]=lg[i>>1]+1;
    scanf("%d",&T);
    while(T--){
        scanf("%d%d",&N,&K); ans=0;
        rep(i,1,N) scanf("%d",&a[i]);
        rep(i,1,N) st[i][0]=i;
        rep(i,1,20) {
            rep(j,1,N+1-(1<<i))
             st[j][i]=a[st[j][i-1]]>=a[st[j+(1<<(i-1))][i-1]]?st[j][i-1]:st[j+(1<<(i-1))][i-1];
        }
    
        rep(i,1,N) vis[i]=0; A[1]=1; vis[a[1]]=1;
        rep(i,2,N){
             if(vis[a[i]])  A[i]=max(A[i-1],vis[a[i]]+1);
             else A[i]=A[i-1];
             vis[a[i]]=i;
        }
        rep(i,1,N) vis[i]=0; B[N]=N; vis[a[N]]=N;
        for(int i=N-1;i>=1;i--){
            if(vis[a[i]]) B[i]=min(B[i+1],vis[a[i]]-1);
            else B[i]=B[i+1];
            vis[a[i]]=i;
        }
        solve(1,N);
        printf("%lld\n",ans);
    }
    return 0;
}