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: Wrong Answer)

import java.io.*
import java.util.*

// N以下で a % b >= Kとなる(a,b)の組
fun solve(N: Int, K: Int){
    if (K == 0){
        println(N*N)
        return
    }
    // a < b
    val ret1 = ((N-K+1)*(N-K))/ 2
    // a > b
    var ret2 = 0
    for (b in K + 1 until N){
        // a = b*x + r <= N -> x = 1,2, ... , (N-r)/b (K<=r<=b-1)
        for (r in K until b){
            ret2 += (N - r)/ b
        }
    }
    println(ret1+ret2)

    return
}


fun main(args: Array<String>) {
    fun StringArray(size: Int, init: (Int) -> String = { "\u0000" }): Array<String> {
        return Array<String>(size, init)
    }
    class Scanner(stream: InputStream) {
        private val reader = BufferedInputStream(stream).bufferedReader()
        private var st: StringTokenizer? = null
        fun next(): String {
            while (!(st?.hasMoreTokens() ?: false)) st = StringTokenizer(reader.readLine())
            return st!!.nextToken()
        }
    }
    val sc = Scanner(System.`in`)
    val N = sc.next().toInt()
    val K = sc.next().toInt()
    solve(N, K)
}