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 <bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;
using vi = vector<int>;
using vvi = vector<vi>;
using vvvi = vector<vvi>;
using vll = vector<ll>;
using vvll = vector<vll>;
using vvvll = vector<vvll>;
using vs = vector<string>;
using pll = pair<ll, ll>;
using vp = vector<pll>;
#define rep(i, n) for(ll i = 0; i < (n); i++)
#define repr(i, a, b) for(ll i = (a); i < (b); i++)
#define ALL(a) (a).begin(), (a).end()
#define SZ(x) ((ll)(x).size())
const ll MOD = 1000000007;
const ll INF = 100000000000000000LL;
inline ll GCD(ll a, ll b){ return b?GCD(b, a % b):a; }
inline ll LCM(ll a, ll b){ return a/GCD(a, b)*b; }
inline ll powint(unsigned long long x, ll y){ ll r=1; while(y){ if(y&1) r*=x; x*=x; y>>=1; } return r; }
inline ll powmod(ll x, ll y, ll m = MOD){ ll r=1; while(y){ if(y&1) r*=x; x*=x; r%=m; x%=m; y>>=1; } return r; }
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; }
#ifdef OJ_LOCAL
#include "dump.hpp"
#else
#define dump(...) ((void)0)
#endif

int main(){
    cin.tie(0); ios::sync_with_stdio(false);
    cout << fixed << setprecision(15);
    ll n, k;
    cin >> n >> k;
    if(k == 0){
        cout << n*n << endl;
        return 0;
    }
    ll ans = 0;
    repr(i, k+1, n+1){
        ans += n/i*(i-k);
        dump(i, ans);
        if(n % i >= k){
        dump(i, ans);
            ans += n-(n/i*i+k)+1;
        }
    }
    cout << ans << endl;
    return 0;
}