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<string>
#include<cstring>
#include<stdio.h>
#include<algorithm>
#include<functional>
#include<vector>
#include<set>
#include<queue>
#include<list>
#include<stdio.h>
#include<stdlib.h>
#include<map>
#include<cstdio>
#include <iomanip>

using namespace std;
typedef long long ll;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;

typedef int _loop_int;
#define REP(i,n) for(_loop_int i=0;i<(_loop_int)(n);++i)
#define FOR(i,a,b) for(_loop_int i=(_loop_int)(a);i<(_loop_int)(b);++i)
#define FORR(i,a,b) for(_loop_int i=(_loop_int)(b)-1;i>=(_loop_int)(a);--i)

#define DEBUG(x) cout<<#x<<": "<<x<<endl
#define DEBUG_VEC(v) cout<<#v<<":";REP(i,v.size())cout<<" "<<v[i];cout<<endl
#define ALL(a) (a).begin(),(a).end()

#define CHMIN(a,b) a=min((a),(b))
#define CHMAX(a,b) a=max((a),(b))

// mod
const ll MOD = 1000000007ll;
#define FIX(a) ((a)%MOD+MOD)%MOD

// floating
typedef double Real;
const Real EPS = 1e-11;
#define EQ0(x) (abs(x)<EPS)
#define EQ(a,b) (abs(a-b)<EPS)


int main() {

	int N, K; cin >> N >> K;

	if (K == 0) {
		cout << N*N << endl;
		return 0;
	}


	int ans = 0;
	for (int b = 1; b < N + 1; b++) {
	
		if (b - 1 >= K) {

			//[nb,nb-1]の区間の数×一個の区間に含まれるK以上の数
			ans += N / b*(b - K);

			//完全な区間にならなかった余り
			ans += max(0, (N%b) - K + 1);
		}

	}

	cout << ans << endl;

	return 0;
}