Heuristic partition

Heuristic split

    given $n$ number, seeking to satisfy certain conditions the number of points or a maximum priority value, and this point of the maximum weight $(a,b)$ section $[a,b]$ interval maximum $/$ Minimum relevant.
    So then you can consider the partition, for the interval $[L,R]$ , find the smallest $/$ Large at the value
    then the processing position across the min / max values of the point where the pair of processing sub section recursively

    For a range, find the location of maximum / minimum values $mid$ can be $RMQ$ pretreatment
    and enumeration $[l, mid]$ 与 $[mid, r]$ smaller interval, the answer to the query processing

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Example a: multi-tenth school field 1011 Make Rounddog Happy

Effect: A good range is defined as $max(a_l, a_{l+1},...,a_r) - (r - l + 1) <= k$
           and the number of all different from each other within the interval, the interval to ask how many good there?
Solution: Divide and Conquer heuristic based on the maximum value, $RMQ$ pretreatment interval maximum value position
            $L[i]$ is to $a[i]$ to the left a number satisfying mutually different farthest position, $R[i]$ to the right, with the double pointer $O (n)$ pretreated to

#include<bits/stdc++.h>
#define rint register int
#define deb(x) cerr<<#x<<" = "<<(x)<<'\n';
using namespace std;
typedef long long ll;
using pii = pair <int,int>;
const int maxn = 3e5 + 5;
int T, n, k, a[maxn];
int L[maxn], R[maxn], vis[maxn];
ll ans;

int dp_max[maxn][20], pos[maxn][20];
void ST(int n, int d[]) {
	for (int i=1; i<=n; i++) {
		dp_max[i][0] = d[i];
		pos[i][0] = i;
	}
	for (int j=1; (1<<j) <= n; j++) {
		for (int i=1; i+(1<<j)-1 <= n; i++) {
			if(dp_max[i][j-1] >= dp_max[i + (1<<(j-1))][j-1]) pos[i][j] = pos[i][j-1];
			else pos[i][j] = pos[i + (1<<(j-1))][j-1];
			dp_max[i][j] = max(dp_max[i][j-1], dp_max[i + (1<<(j-1))][j-1]);
		}
	}
}
int RMQ_max(int l, int r) {
	int k = 0;
	while ((1<<(k+1)) <= r-l+1) k++;
	if(dp_max[l][k] >= dp_max[r - (1<<k)+1][k]) return pos[l][k];
	else return pos[r - (1<<k)+1][k];
}

void solve(int l, int r) {
	if(l > r) return;
	int mid = RMQ_max(l, r);
	if(mid - l <= r - mid) {
		for(int i=l; i<=mid; i++) {
			int r1 = max(mid, a[mid] - k + i - 1);
			int r2 = min(r, R[i]);
			if(r1 <= r2) ans += r2 - r1 + 1;
		}
	} else {
		for(int i=mid; i<=r; i++){
			int l2 = min(mid, -a[mid] + k + i + 1);
			int l1 = max(l, L[i]);
			if(l1 <= l2) ans += l2 - l1 + 1;
		}
	}
	solve(l, mid - 1), solve(mid + 1, r);
}

int main() {
	scanf("%d", &T);
	while(T--) {
		scanf("%d%d", &n, &k);
		for(int i=1; i<=n; i++) scanf("%d", a+i);
		vis[a[1]] = 1;
		for(int i=1, p=2; i<=n; i++) {
			while(p<=n && !vis[a[p]]) vis[a[p++]] = 1;
			vis[a[i]] = 0, R[i] = p - 1;
		}
		vis[a[n]] = 1;
		for(int i=n, p=n-1; i; i--) {
			while(p && !vis[a[p]]) vis[a[p--]] = 1;
			vis[a[i]] = 0, L[i] = p + 1;
		}
		ans = 0;
		ST(n, a);
		solve(1, n);
		printf("%lld\n", ans);
	}
}

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Example two: Luo Gu P4755 Beautiful Pair

Effect: asked how many number of $a_i, a_j (i<=j)，a_i \times a_j <= max(a_i, a_{i+1},...,a_j)$
Solution: heuristic based on the maximum division, enumeration $a_i$To find out how many sections $<=\frac{mx}{a_i}$ The number can be used to Chairman tree

#include<bits/stdc++.h>
#define rint register int
#define deb(x) cerr<<#x<<" = "<<(x)<<'\n';
using namespace std;
typedef long long ll;
const int maxn = 2e5 + 5;
int n, a[maxn], aid[maxn], tot, t[maxn*30]; 
int root[maxn*30], ls[maxn*30], rs[maxn*30];
vector <int> v;
ll ans;

int dp_max[maxn][20], pos[maxn][20];
void ST(int n, int d[]) {
	for (int i=1; i<=n; i++) {
		dp_max[i][0] = d[i];
		pos[i][0] = i;
	}
	for (int j=1; (1<<j) <= n; j++) {
		for (int i=1; i+(1<<j)-1 <= n; i++) {
			if(dp_max[i][j-1] >= dp_max[i + (1<<(j-1))][j-1]) pos[i][j] = pos[i][j-1];
			else pos[i][j] = pos[i + (1<<(j-1))][j-1];
			dp_max[i][j] = max(dp_max[i][j-1], dp_max[i + (1<<(j-1))][j-1]);
		}
	}
}
int RMQ_max(int l, int r) {
	int k = 0;
	while ((1<<(k+1)) <= r-l+1) k++;
	if(dp_max[l][k] >= dp_max[r - (1<<k)+1][k]) return pos[l][k];
	else return pos[r - (1<<k)+1][k];
}

int getid(int x){
	return lower_bound(v.begin(), v.end(), x) - v.begin() + 1;
}

void update(int &rt, int pre, int l, int r, int pos){
	if(l>pos || r<pos) return;
	rt = ++tot;
	t[rt] = t[pre] + 1;
	ls[rt] = ls[pre], rs[rt] = rs[pre];
	if(l == r) return;
	int m = l + r >> 1;
	update(ls[rt], ls[pre], l, m, pos);
	update(rs[rt], rs[pre], m+1, r, pos);
}

int query(int rt, int pre, int l, int r, int pos){
	if(l > pos) return 0;
	if(r <= pos) return t[rt] - t[pre];
	int m = l + r >> 1, ret = 0;
	ret += query(ls[rt], ls[pre], l, m, pos);
	ret += query(rs[rt], rs[pre], m+1, r, pos);
	return ret;
}

void solve(int l, int r){
	if(l > r) return;
	int pos = RMQ_max(l, r), mx = a[pos];
	if(pos - l <= r - pos){
		for(int i=l; i<=pos; i++){
			int k = mx / a[i];
			int kid = upper_bound(v.begin(), v.end(), k) - v.begin();
			ans += query(root[r], root[pos-1], 1, v.size(), kid);
		}
	} else {
		for(int i=pos; i<=r; i++){
			int k = mx / a[i];
			int kid = upper_bound(v.begin(), v.end(), k) - v.begin();
			ans += query(root[pos], root[l-1], 1, v.size(), kid);
		}
	}
	solve(l, pos - 1), solve(pos + 1, r);
}

int main() {
	scanf("%d", &n);
	for(int i=1; i<=n; i++) scanf("%d", a+i), v.push_back(a[i]);
	v.push_back(1e9 + 7);
	sort(v.begin(), v.end());
	v.erase(unique(v.begin(), v.end()), v.end());
	for(int i=1; i<=n; i++){
		aid[i] = getid(a[i]);
		update(root[i], root[i-1], 1, v.size(), aid[i]);
	}
	ST(n, a);
	solve(1, n);
	printf("%lld\n", ans);
}