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 Kotlin (1.0.0) (judge verdict: Accepted)

fun main(args: Array<String>) {
    val (n, k) = readLine()!!.split(" ").map { it.toLong() }

    // a, b <= n && a % b >= k

    fun 特定の割る数に対してすべての余りを考慮できる場合(divisor: Long): Long {

        var patternNum = n / divisor
        var restSetNum = divisor - k

        return patternNum * restSetNum
    }

    fun 特定の割る数に対して上限Nの関係で一部の余りしか考慮できない場合(divisor: Long): Long {

        var num: Long

        if (k != 0L) {
            num = n % divisor - (k - 1)
        } else {
            num = n % divisor
        }

        return if (num < 0) 0 else num
    }

    var count = 0L

    for (i in (k + 1)..n) {
        // 割る数を k ~ n に変えていったときの数を算出
        // 最後の一つ手前まで + 最後余ったところ
        count += 特定の割る数に対してすべての余りを考慮できる場合(i) + 特定の割る数に対して上限Nの関係で一部の余りしか考慮できない場合(i)
    }

    print(count)
}

// 考慮するべきケース
// k = 0 の場合 -> OK
// n = 1 の場合 -> OK
// 結果が Int に収まらないケース !!