Problem:
Score : 400 points

Problem Statement

Takahashi had a pair of two positive integers not exceeding N, (a,b), which he has forgotten.
He remembers that the remainder of a divided by b was greater than or equal to K.
Find the number of possible pairs that he may have had.

Constraints

1 \leq N \leq 10^5

0 \leq K \leq N-1

All input values are integers.

Input

Input is given from Standard Input in the following format:

N K

Output

Print the number of possible pairs that he may have had.

Sample Input 1

5 2

Sample Output 1

7

There are seven possible pairs: (2,3),(5,3),(2,4),(3,4),(2,5),(3,5) and (4,5).

Sample Input 2

10 0

Sample Output 2

100

Sample Input 3

31415 9265

Sample Output 3

287927211

Solution in Rust (1.15.1) (judge verdict: Accepted)

#[allow(unused_imports)]
use std::io::*;
#[allow(unused_imports)]
use std::str::*;
#[allow(unused_imports)]
use std::mem::*;
#[allow(unused_imports)]
use std::cmp::*;
#[allow(unused_imports)]
use std::collections::HashMap;
#[allow(unused_imports)]
use std::collections::VecDeque;
#[allow(unused_imports)]
use std::usize;

#[allow(unused_macros)]
macro_rules! read_cols {
    ($($t:ty),+) => {{
        let mut line = String::new();
        stdin().read_line(&mut line).unwrap();

        let mut it = line.trim()
            .split_whitespace();

        ($(
            it.next().unwrap().parse::<$t>().ok().unwrap()
        ),+)
    }}
}

#[allow(dead_code)]
fn read<T: FromStr>() -> T {
    let mut line = String::new();
    stdin().read_line(&mut line).unwrap();
    line.trim().to_string().parse().ok().unwrap()
}

#[allow(dead_code)]
fn read_vec<T: FromStr>() -> Vec<T> {
    let mut line = String::new();
    stdin().read_line(&mut line).unwrap();

    line.trim()
        .split_whitespace()
        .map(|s| s.parse().ok().unwrap())
        .collect()
}

fn main() {
    let (n, k) = read_cols!(i64, i64);

    let mut ans = 0;
    for b in k + 1..n + 1 {
        // b以下のa候補(0含む) => b - k
        // これらにbを足したものも候補になる
        // => (b - k) * [n / b] + 半端分
        ans += (b - k) * (n / b);

        if (n / b) * b + k <= n {
            ans += n - ((n / b) * b + k) + 1;
        }

        if k == 0 {
            // 0の分を引く
            ans -= 1;
        }
    }

    println!("{}", ans);
}