Problem 1

Description

Find the number of all intervals in an array with mean greater than or equal to t

Solution

sumisum_i denotes the prefix sum of the array from 1 to i,

then the mean of interval [i+1,j][i+1,j] is

sumj−sumij−i\frac{sum_j - sum_i}{j-i}
sumj−sumij−i≥t\frac{sum_j - sum_i}{j-i} \geq t
sumj−t×j≥sumi−t×isum_j - t \times j \geq sum_i - t \times i

Code 1

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#include <iostream>
#include <algorithm>
#define ll long long
using namespace std;

const int MAXN = 1e6 + 10;
const ll MOD = 1e9 + 7;
ll n, t;
ll a[MAXN], b[MAXN], c[MAXN];
ll bit[MAXN];

ll lowbit(ll i) {
    return i & (-i);
}

void add(ll i, int x) {
    for (; i <= n + 1; i += lowbit(i))
        bit[i] += x;
}

ll sum(ll i) {
    ll ans = 0;
    for (; i; i -= lowbit(i))
        ans += bit[i];
    return ans;
}

int main() {
    cin >> n >> t;
    for (int i = 1; i <= n; i++) {
        cin >> a[i];
        a[i] += a[i - 1];
    }
    for (int i = 1; i <= n; i++) {
        b[i] = a[i] - i * t;
        c[i] = b[i];
    }
    sort(c, c + 1 + n);
    ll ans = 0;
    for (int i = 0; i <= n; i++) {
        ll pos = lower_bound(c, c + 1 + n, b[i]) - c + 1;
        ans = (ans + sum(pos)) % MOD;
        add(pos, 1);
    }
    cout << (ll) ans << endl;
}

Code 2

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#include <algorithm>
#include <iostream>
#define ll long long
using namespace std;

const int MAXN = 1e6 + 10;
const ll MOD = 1e9 + 7;
ll n, t, ans;
ll a[MAXN];
ll q[MAXN];

void merge_sort(int l, int r, ll *a) {
    if (l >= r)
        return;
    int mid = l + r >> 1;
    merge_sort(l, mid, a);
    merge_sort(mid + 1, r, a);
    int i = l, j = mid + 1, t = 0;
    while (i <= mid && j <= r) {
        if (a[i] > a[j]) {
            q[t++] = a[j++];
            ans += mid - i + 1;
            ans %= MOD;
        } else
            q[t++] = a[i++];
    }
    while (i <= mid) q[t++] = a[i++];
    while (j <= r) q[t++] = a[j++];
    for (int i = l, j = 0; i <= r; i++, j++)
        a[i] = q[j];
}

int main() {
    cin >> n >> t;
    for (int i = 1; i <= n; i++) {
        cin >> a[i];
        a[i] += a[i - 1] - t;
    }
    merge_sort(0, n, a);
    cout << (ll) (n * (n + 1) / 2 - ans) % MOD << endl;
    return 0;
}

Problem 2

Description

Maximize the median sweetness level of m desserts while staying within a budget of v.

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#include <algorithm>
#include <iostream>
#include <queue>
using namespace std;
const int MAX_N = 1e5 + 10;
struct sweet {
    int ness;//sweetness
    int price;
} sweets[MAX_N];
bool cmp(sweet x, sweet y) { return x.ness < y.ness; }
int v, n, m;
int sum1[MAX_N], sum2[MAX_N];
int main() {
    int ans = 0;
    cin >> v >> n >> m;
    for (int i = 1; i <= n; i++)
        cin >> sweets[i].ness >> sweets[i].price;
    sort(sweets + 1, sweets + 1 + n, cmp);
    priority_queue<int> q1, q2;
    // compute the sum of prices from the left end of the array
    for (int i = 1; i <= n; i++) {
        q1.push(sweets[i].price);
        sum1[i] = sum1[i - 1] + sweets[i].price;
        if (q1.size() > m / 2 - 1 + m % 2)
            sum1[i] -= q1.top(), q1.pop();
    }
    // compute the sum of prices from the right end of the array
    for (int i = n; i > 0; i--) {
    for (int i = n; i > 0; i--) {
        q2.push(sweets[i].price);
        sum2[i] = sum2[i + 1] + sweets[i].price;
        if (q2.size() > m / 2)
            sum2[i] -= q2.top(), q2.pop();
    }
    if (m % 2) {
        // search for the sweet with the maximum sweetness that satisfies the budget constraint
        for (int i = n - m / 2; i >= m / 2 + 1; i--)
            if (sum1[i - 1] + sum2[i + 1] + sweets[i].price <= v) {
                cout << sweets[i].ness;
                break;
            }
    } else {
        for (int i = m / 2; i <= n - m / 2; i++) {
            int l = i + 1, r = n - m / 2 + 1;
            int tmp = 0;
            while (l <= r) {
                int mid = l + (r - l) / 2;
                if (sum1[i - 1] + sum2[mid] + sweets[i].price <= v)
                    l = mid + 1, tmp = mid;
                else
                    r = mid - 1;
            }
            if (tmp > i)
                ans = max(ans, sweets[i].ness + sweets[tmp].ness);
        }
        cout << ans / 2 << "\n";
    }
    return 0;
}