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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

/*------------------------------------
........Bismillahir Rahmanir Rahim....
..........created by Abdul Aziz.......
------------------------------------*/
#include <iostream>
#include <algorithm>
#include <stdio.h>
#include <cmath>
#include <vector>
#include <set>
#include <list>
#include <stack>
#include <map>
#include <cstring>
#include <unordered_map>
#include <queue>
#define   mod  998244353
#define   int  long long 
#define   ld   long double
#define   pb   push_back
#define   vi  vector<int> 
#define   dbg(x)  cerr << #x << " = " << x << '\n'
#define   sz(x)  (int)x.size()
#define   all(a) (a.begin(),a.end())
#define   ff   first
#define   ss   second
#define   pii  pair<int,int> 
#define   lcm(a,b) (a*b)/__gcd(a,b) 
using namespace std;

inline void solve(){
    int  n, k, ans = 0; 
    cin >> n >> k ;
    for (int b=1;b<=n;b++ ) {
        ans += n/b*max(0LL,b-k); 
        ans += max(0LL,(n%b)-k+1);
    } 
   // if (k==0) ans = n*n;
    cout << ans << endl;
}

signed main()
{
    int n=1; // cin>>n;
    while (n--) solve();
    return 0;
}