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补题链接：

超级基本的二分

贴个代码就溜了

using ll = long long;void solve() {    ll n, k;    cin >> n >> k;    vector
   
     a(n);    for (ll &x : a) cin >> x;    ll l = 1, r = 1e9, len;    while (l <= r) {        ll mid = l + r >> 1, sum = 0;        for (int i = 0; i < n; ++i) sum += a[i] / mid;        if (sum >= k) len = mid, l = mid + 1;        else            r = mid - 1;    }    cout << len << '\n';}