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 C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <iostream>
#include <math.h>
#include <algorithm>
#include <vector>
#include <numeric>
#include <string>
using namespace std;
const double PI = acos(-1.0);
const string alp = "abcdefghijklmnopqrstuvwxyz";
const string ALP = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
#define EXIST(s,e) ((s).find(e)!=(s).end())
#define SORT(c) sort((c).begin(),(c).end())
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n) FOR(i,0,n)
#define REP1(i,n) FOR(i,1,n+1)
#define SUM(c) accumulate((c).begin(),(c).end(),0)
#define UNIQ(c) (c).erase(unique((c).begin(), (c).end()), (c).end());
#define Yn(b) puts(b?"Yes":"No")
#define YN(b) puts(b?"YES":"NO")

bool isprime(int p) {
  if (p==1) return false;
    FOR(i,2,p) {
      if (p%i==0) return false;
    }
  return true;
}

int main(){
  int n,k,c;
  cin >> n >> k;
  c=0;
  if(k==0){
    cout << n*n;
    return 0;
  }
  FOR(i,k,n+1){
    c+=(n/i)*(i-k);
    if((n%i)-k+1>0){
      c+=(n%i)-k+1;
    }
  }
  cout << c;
  return 0;
}