K non-decimal query evaluation practices k decimal Chairman tree

Topic Link
title meaning
given a length n number of columns A, there are m times the next operation, there are two operations:
. 1 LRX, expressed i belonging to [l, r], = min so Ai (Ai, X)
2 LRK , indicating an inquiry interval [l, r] in a smaller number of k.
Thinking
small number of k, that is less than the k-k-1 there is a small one, consider adding half block, find the time to ask block is less than the number of how many mid, approaching half the answer.

#include <bits/stdc++.h>
#define INF 0x3f3f3f3f
using namespace std;
const int maxn = 1e5 + 7;
typedef long long ll;
int belong[maxn], l[2005], r[2005], tag[2005], num, block, n, q;
vector<int> v[2005];
int a[maxn];
void build()
{
    block = sqrt(n);
    num = n / block;
    if(n % block) num++;
    for (int i = 1; i <= num; i++)
        l[i] = (i - 1) * block + 1, r[i] = i * block;
    r[num] = n;
    for (int i = 1; i <= n; i++) {
        belong[i] = (i - 1) / block + 1;
        v[belong[i]].push_back(a[i]);
    }
    for (int i = 1; i <= num; i++) {
        tag[i] = INF;
        sort(v[i].begin(), v[i].end());
    }
}
void reset(int x)
{
    v[x].clear();
    for (int i = l[x]; i <= r[x]; i++) {
        a[i] = min(a[i], tag[x]);
        v[x].push_back(a[i]);
    }
    sort(v[x].begin(), v[x].end());
}
void update(int x, int y, int c) {
    int tl = belong[x], tr = belong[y];
    if(tl == tr) {
        for (int i = x; i <= y; i++)
            a[i] = min(a[i], c);
        reset(tl);
        return ;
    }
    for (int i = x; i <= r[tl]; i++)
        a[i] = min(a[i], c);
    reset(tl);
    for (int i = l[tr]; i <= y; i++)
        a[i] = min(a[i], c);
    reset(tr);
    for (int i = tl + 1; i < tr; i++)
        tag[i] = min(tag[i], c);
}
int query(int x, int y, int c)
{
    int tl = belong[x], tr = belong[y], ans = 0;
    if(tl == tr) {
        for (int i = x; i <= y; i++) {
            if(a[i] <= c || tag[tl] <= c) ans++;
        }
        return ans;
    }
    for (int i = x; i <= r[tl]; i++) {
        if(a[i] <= c || tag[tl] <= c) ans++;
    }
    for (int i = l[tr]; i <= y; i++) {
        if(a[i] <= c || tag[tr] <= c) ans++;
    }
    for (int i = tl + 1; i < tr; i++) {
        if(tag[i] <= c) ans += (r[i] - l[i] + 1);
        else ans += upper_bound(v[i].begin(), v[i].end(), c) - v[i].begin();
    }
    return ans;
}
int check(int x, int y, int k) {
    int left = -1e9, right = 1e9;
    int ans;
    while (left <= right) {
        int mid = (left + right) / 2;
        if(query(x, y, mid) >= k) {
            ans = mid;
            right = mid - 1;
        }
        else left = mid + 1;
    }
    return ans;
}
int main()
{
    scanf("%d%d", &n, &q);
    for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
    build();
    for (int i = 1; i <= q; i++) {
        int op, x, y, z;
        scanf("%d%d%d%d", &op, &x, &y, &z);
        if(op == 1) update(x, y, z);
        else printf("%d\n", check(x, y, z));
    }
}

Multiplication
problem intended
to A, B two episodes, seeking a larger value of K is ai * bi.
Ideas
seeking the K a big problem, with half the number of inquiries is larger than the mid how many.

#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 7;
typedef long long ll;
ll a[maxn], b[maxn], k;
int n, m;
bool check(ll val) {
    ll cnt = 0;
    for (int i = 1; i <= n; i++) {
        if(a[i] == 0) cnt += val < 0 ? m : 0;
        else if(a[i] < 0) cnt += lower_bound(b + 1, b + 1 + m, ceil(1.0*val/a[i])) - b - 1;
        else cnt += (m - (upper_bound(b + 1, b + 1 + m, floor(1.0 * val / a[i])) - b - 1));
    }
    return cnt <= k;
}
int main()
{
    scanf("%d%d%lld", &n, &m, &k);
    k--;
    for (int i = 1; i <= n; i++) scanf("%lld", &a[i]);
    for (int i = 1; i <= m; i++) scanf("%lld", &b[i]);
    sort(a + 1, a + 1 + n); sort(b + 1, b + 1 + m);
    ll left = -1e12, right = 1e12, ans;
    while (left <= right) {
        ll mid = (left + right) / 2;
        if(check(mid)) {
            ans = mid;
            right = mid - 1;
        }
        else left = mid + 1;
    }
    printf("%lld\n", ans);
}