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;

#define DUMP(x) std::cerr << (#x) << " = " << (x) << "\n"
#define rep(i, n) for (int i = 0; i < (int)(n); ++i)
#define REP(i, k, n) for (int i = (k); i < (int)(n); ++i)
#define ALL(r) r.begin(), r.end()
#define YES puts("YES")
#define Yes puts("Yes")
#define NO puts("NO")
#define No puts("No")
#define IMP puts("IMPOSSIBLE")
#define Imp puts("Impossible")
#define imp puts("impossible")
#define M1 puts("-1")
#define pb push_back

template <typename T>
T dup(T x, T y)
{
    return (x + y - 1) / y;
};
template <typename A, size_t N, typename T>
inline void arrayFill(A (&array)[N], const T &val) { std::fill((T *)array, (T *)(array + N), val); }

using ll = int64_t;
using vint = vector<int32_t>;
using vvint = vector<vint>;
using vll = vector<ll>;
using vvll = vector<vll>;
using vstr = vector<string>;
using pint = pair<int32_t, int32_t>;
using pll = pair<ll, ll>;
using setint = set<int32_t>;
using setstr = set<string>;
using qint = queue<int32_t>;
using qpint = queue<pint>;

constexpr std::int32_t INF = 1001001001;
constexpr std::int64_t LINF = 1001001001001001001;

void Main()
{
    ll n, k;
    cin >> n >> k;

    if (k == 0)
    {
        cout << n * n << endl;
        return;
    }

    ll ans = 0;
    ll cnt = 0;
    rep(b, n + 1)
    {
        if (b <= k)
            continue;
        ++cnt;
        ans += ((n + 1) / b * cnt) + max(0l, (n + 1) % b - k);
    }
    cout << ans << endl;
}

int main()
{
    std::cin.tie(nullptr);
    std::ios_base::sync_with_stdio(false);
    std::cout << std::fixed << std::setprecision(15);
    Main();
    return 0;
}