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: Accepted)

#include <algorithm>
#include <cmath>
#include <deque>
#include <iostream>
#include <numeric>
#include <sstream>
#include <string>
#include <string.h>
#include <tuple>
#include <vector>


#define REP(i,x) for(int i{ 0 }; i < (int)(x); i++)
#define REPC(i,x) for(int i{ 0 }; i <= (int)(x); i++)
#define RREP(i,x) for(int i{ (int)(x) - 1 }; i >= 0 ;i--)
#define RREPC(i,x) for(int i{ (int)(x)}; i >= 0; i--)
#define REP1O(i,x) for(int i{ 1 }; i < (int)(x); i++)
#define REP1C(i,x) for(int i{ 1 }; i <= (int)(x); i++)

#define PB push_back
#define MP make_pair
#define F first
#define S second
#define SZ(x) ((int)(x).size())
#define ALL(x) (x).begin(),(x).end()

using namespace std;

typedef int64_t ll;
typedef pair<int, int> pii;
typedef tuple<int, int, int> tupiii;

const int dx[4] = { 1, 0, -1,  0 };
const int dy[4] = { 0, 1,  0, -1 };
const int INTMAX = 2147483647;
const ll LLMAX = 9223372036854775807;

template<class T>bool chmax(T& a, const T& b) { if (a < b) { a = b; return 1; } return 0; }
template<class T>bool chmin(T& a, const T& b) { if (b < a) { a = b; return 1; } return 0; }
void SWAP(ll& a, ll& b) { a ^= b; b ^= a; a ^= b; }
void SWAP(int& a, int& b) { a ^= b; b ^= a; a ^= b; }


class D {
	int n, k;
public:
	D()
	{
		cin >> n >> k;
	}
	void solve()
	{
		ll ans{ 0 };
		for (int div{ k + 1}; div <= n; div++)
		{
			int loops{n / div};
			ans += ((ll)div - 1 - (k - 1)) * loops;
			ans += (ll)max(0, n - (div * loops + k) + 1);
		}

		if (k == 0)
		{
			ans -= (ll)n;
		}

		cout << ans << endl;
	}
};


int main()
{
	D solution;
	solution.solve();

	return 0;
}