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You will be given an array and some queries. Each query will contain three integers l, r, and x. You have to return the count of the integer in array in the range [l,r] whose value is less than or equal to x. for example n=7 q=2 5 4 8 7 6 2 1 1 3 5 2 5 6

ans = 2 2

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
#define ll long long
#define int long long
#define endl "\n"
#define ptr(a, n)               \
    for (int i = 0; i < n; i++) \
    {                           \
        cout << a[i] << ' ';    \
    }
#define fort(z, x, t) for (ll z = x; z < t; z++)
#define rev(z, x, t) for (ll z = t; z > x; z--)
#define arin(a, n)             \
    ll a[n];                   \
    for (ll i = 0; i < n; i++) \
    {                          \
        cin >> a[i];           \
    }
#define pb push_back
#define vein(v, n)             \
    vector<ll> v;              \
    for (ll i = 0; i < n; i++) \
    {                          \
        ll x;                  \
        cin >> x;              \
        v.pb(x);               \
    }
#define seint(s, n)            \
    set<int> s;                \
    for (ll i = 0; i < n; i++) \
    {                          \
        ll x;                  \
        cin >> x;              \
        s.insert(x);           \
    }
#define vint(v) vector<ll> v;
#define vstring(vs, n)         \
    vector<string> vs;         \
    for (ll i = 0; i < n; i++) \
    {                          \
        string s;              \
        cin >> s;              \
        vs.pb(s);              \
    }
#define num(n) \
    ll n;      \
    cin >> n;
#define out(n) cout << n << endl;
#define sortv(v) sort(v.begin(), v.end());
#define mpll map<ll, ll> mpi;
#define mpchar map<char, ll> mpc;
#define cin(a, n, b) \
    f(i, n)          \
    {                \
        cin >> a[i]; \
        b[i] = a[i]; \
    }
#define yes cout << "YES" << endl;
#define no cout << "NO" << endl;
typedef tree<ll, null_type, less_equal<ll>, rb_tree_tag, tree_order_statistics_node_update> ordered_multiset;
typedef tree<ll, null_type, less<ll>, rb_tree_tag, tree_order_statistics_node_update> ordered_set;
const int N = 1e5 + 10;
int mod = 1e9 + 7;
int fun(int mid, int k, vector<int> &v)
{
    int n = v.size();
    int val = 0;
    for (int i = 0; i < n; i++)
    {
        val += max(0ll, mid - v[i]);
        if (val > k)
            return 0;
    }
    if (val > k)
        return 0;
    return 1;
}
void solve()
{
    num(n);
    num(k);
    vein(v, n);
    sortv(v);
    int req = 0;
    for (int i = 0; i < n; i++)
    {
        req += max(0ll, v.back() - v[i]);
    }
    if (req <= k)
    {
        k -= req;
        int val1 = k / n;
        int val2 = k % n;
        for (int i = 0; i < n; i++)
        {
            v[i] = val1 + v.back();
        }
        sortv(v);
        for (int i = 0; i < val2; i++)
            v[i]++;
        vector<int> suf(n + 1, 0);
        suf[n - 1] = v[n - 1];
        for (int i = n - 2; i >= 0; i--)
        {
            suf[i] = suf[i + 1] + v[i];
        }
        int ans = 0;
        for (int i = 0; i < n - 1; i++)
        {
            ans += (v[i] * suf[i + 1]) % mod;
            ans %= mod;
        }
        out(ans % mod);
        return;
    }
    sortv(v);
    int miin = v[0];
    int maax = v.back();
    while (miin + 1 < maax)
    {
        int mid = (miin + maax) / 2;
        if (fun(mid, k, v))
        {
            miin = mid;
        }
        else
        {
            maax = mid;
        }
    }
    for (int i = 0; i < n; i++)
    {
        if (v[i] < miin)
        {
            int val2 = min(k, miin - v[i]);
            v[i] += val2;
            k -= val2;
        }
    }
    sortv(v);
    for (int i = 0; i < k; i++)
    {
        v[i]++;
    }
    sortv(v);
    vector<int> suf(n + 1, 0);
    suf[n - 1] = v[n - 1];
    for (int i = n - 2; i >= 0; i--)
    {
        suf[i] = suf[i + 1] + v[i];
    }
    int ans = 0;
    for (int i = 0; i < n - 1; i++)
    {
        ans += (v[i] * suf[i + 1]) % mod;
        ans %= mod;
    }
    // ptr(v, n);
    // cout << endl;
    out(ans % mod);
}
signed main()
{
    // #ifndef ONLINE_JUDGE
    // freopen("input.txt", "r", stdin);
    // freopen("output.txt", "w", stdout);
    // #endif
    std::ios::sync_with_stdio(false);
    std::cin.tie(0);
    int t;
    cin >> t;
    fort(i, 0, t)
        solve();
    return 0;
}

Please help me, I don't know what is wrong. Ploblem Link

I solved the problem by taking the maximum weight first and then further checking if one more child can be fit into the same gondola. As per the question, atmost 2 children can only sit in one gondola but I was thinking if this constraint was not present then will the same solution work i.e first try to make sit all the heavy weights children till weight does not exceed x and then add new children of small weights till we did not exceed the limit. I am confused will the same solution will work or not Question link