整除（数论）分块

$\lfloor \frac{n}{l} \rfloor = \lfloor \frac{n}{l + 1} \rfloor = ... = \lfloor \frac{n}{r} \rfloor$
$\lfloor \frac{n}{l} \rfloor \le \frac{n}{r} < \lfloor \frac{n}{l} \rfloor + 1$
根据不等式左侧，得到 $r \le \lfloor \frac{n}{\lfloor \frac{n}{l} \rfloor} \rfloor$

例题：https://www.luogu.com.cn/problem/UVA11526

#include<bits/stdc++.h>
using namespace std;
using LL = long long;
void solve(){
	LL n;
	cin >> n;
	LL ans = 0;
	for (LL i = 1, j; i <= n; i = j + 1){
		j = n / (n / i);
		ans += (LL)(j - i + 1) * (n / i);
	}
	cout << ans << "\n";
}
int main(){
	ios::sync_with_stdio(false);cin.tie(0);
	int T;
	cin >> T;
	while (T -- ){
		solve();
	}
	return 0;
}