题解 | 小红的矩阵

#include <iostream>
using namespace std;
using ll = long long;

ll n, m, k;

ll count(ll x) {
    ll cnt = 0;
    for (ll i = 1; i <= n; i++) {
        cnt += min(m, x / i);
    }
    return cnt;
}

int main() {
    cin >> n >> m >> k;
    ll l = 1, r = n * m;

    while (l < r) {
        ll mid = (l + r) / 2;

        if (count(mid) >= k) r = mid;
        else l = mid + 1;
    }

    cout << l << endl;
    return 0;
}