lc 1862

class Solution
{
    using ll = long long;
    static const int MOD = 1e9 + 7;

    ll get_block_cnt(int i, const vector<int> &v)
    {
        // ll start = v[i];
        ll ret = 0;
        ll n = v.back();

        for (ll l = v[i]; l <= n; l++) {
            ll d = n / l, r = n / d;
            int lb = lower_bound(v.begin(), v.end(), l) - v.begin();
            int ub = upper_bound(v.begin(), v.end(), r) - v.begin() - 1;
            if (i == 0) {
                printf("lb:%d ub:%d d:%lld l:%lld, r:%lld\n", lb, ub, d, l, r);
            }
            ret += (ub - lb + 1) * d;
            l = r;
        }
        printf("i:%d ret:%lld\n", i, ret);
        return ret;
    }

  public:
    int sumOfFlooredPairs(vector<int> &ns)
    {
        int n = ns.size();
        sort(ns.begin(), ns.end());
        ll ret = 0;
        for (int i = 0; i < n; ++i) {
            ret += get_block_cnt(i, ns);
        }
        return ret;
    }
};