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)

use std::io::*;
use std::str::FromStr;

pub fn read<T: FromStr>() -> T {
    let stdin = stdin();
    let stdin = stdin.lock();
    let token: String = stdin
        .bytes()
        .map(|c| c.expect("failed to read char") as char)
        .skip_while(|c| c.is_whitespace())
        .take_while(|c| !c.is_whitespace())
        .collect();
    token.parse().ok().expect("failed to parse token")
}

use std::cmp::max;

fn main() {
    let n: usize = read();
    let k: i64 = read();
    let mut ans: i64 = 0;

    for i in (k + 1) as usize..n + 1 {
        // println!("{} % {} = {} ... {}", n, i, n / i, n % i);
        if n / i != 0 {
            ans += max((n / i) as i64 * (i as i64 - k), 0);
        }
        // println!("{}", ans);
        if n % i != 0 {
            if k == 0 {
                ans += max((n % i) as i64 - k, 0);
            } else {
                ans += max((n % i) as i64 - k + 1, 0);
            }
        }
        // println!("{}", ans);
    }
    println!("{}", ans);
}