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 <iostream>
#include <stack>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <map>
#include <set>
#include <tuple>
#include <algorithm>
#include <functional>
#include <cstring>
#include <limits.h>
#include <bitset>
#include <complex>
#include <cassert>
#define FOR(i,k,n)  for (int i=(k); i<(int)(n); ++i)
#define REP(i,n)    FOR(i,0,n)
#define FORIT(i,c)	for(__typeof((c).begin())i=(c).begin();i!=(c).end();++i)
#define SZ(i) ((int)i.size())
#define GI(i) (scanf("%d",&i))
#define GLL(i) (scanf("%lld",&i))
#define GD(i)  (scanf("%lf",&i))
#define PB          push_back
#define MP          make_pair
#define MT          make_tuple
#define GET0(x)     (get<0>(x))
#define GET1(x)     (get<1>(x))
#define GET2(x)     (get<2>(x))
#define ALL(X)      (X).begin(),(X).end()
#define LLMAX       (1LL<<60)
#define LLMIN       -(1LL<<60)
#define IMAX        (1<<30)
#define IMIN        -(1<<30)
typedef long long LL;
using namespace std;

int main(void){
    LL N,K;
    GLL(N);GLL(K);
    if(K==0){
        printf("%lld\n",N*N);
        return 0;
    }
    LL ans = 0;
    for(LL b = K+1;b<=N;b++){
        // printf("b=%lld\n",b);
        for(LL a = 0;a<=N;a+=b){
            // printf("a=%lld\n",a);
            // printf("min(N,a+b-1) - a+(K-1) = %lld\n",min(N,a+b-1) - (a+(K-1)));
            ans += max((LL)0,min(N,a+b-1) - (a+(K-1)));
        }
    }
    printf("%lld\n",ans);
    return 0;
}