考慮每個數的貢獻，即滿足區間第k大為該數的區間個數。其實就是對於兩邊比起大的數，兩邊卡出比該數大的個數為k - 1.
用連結串列維護，考慮從小到大列舉，列舉完成後刪除該點,保證每次列舉連結串列中都是比其大的數。
O(n*k)

寫的挫

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const int MAXN = 5e5 + 10;
typedef long long ll;
struct Node
{
	int next, pre;
	int pos;
	Node(){
		next = -1;
		pre = -1;
	}
}ls[MAXN];
int n, a[MAXN], k;
void init()
{
	for (int i = 0; i < MAXN; i++){
		ls[i].pre = ls[i].next = -1;
	}
}
int main()
{
	int T;
	scanf("%d", &T);
	while (T--){
		init();
		scanf("%d%d", &n, &k);
		a[0] = 0;
		a[n + 1] = n + 1;
		for (int i = 1; i <= n; i++){
			scanf("%d", &a[i]);
		}
		for (int i = 1; i <= n; i++){
			ls[a[i]].pre = a[i - 1];
			ls[a[i]].next = a[i + 1];
			ls[a[i]].pos = i;
		}
		ls[0].next = a[1];
		ls[0].pos = 0;
		ls[n + 1].pos = n + 1;
		ls[n + 1].pre = a[n];
		ll ans = 0;
		for (int i = 1; i <= n; i++){
			int low = i;
			int left = 0;
			while (true){
				if (left == k)break;
				if (low == 0)break;
				low = ls[low].pre;
				left++;
			}
			int right = 0;
			int up = i;
			while (true){
				if (right == k - left)break;
				if (up == n + 1)break;
				up = ls[up].next;
				right++;
			}
			while (low != i&&up != n + 1){		//滑動
				ans += (ll)(ls[ls[low].next].pos - ls[low].pos )
					*(ll)(ls[ls[up].next].pos - ls[up].pos) * (ll)i;
				low = ls[low].next;
				up = ls[up].next;
			}
			ls[ls[i].next].pre = ls[i].pre;		//刪點
			ls[ls[i].pre].next = ls[i].next; 
		}
		printf("%lld\n", ans);
	}
	return 0;
}