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)




/*
    AUTHOR:hruday pabbisetty
    NIT ROURKElA
    I am too dumb to realize that i am too dumb.
*/
#include <bits/stdc++.h>
#include <tr1/unordered_map>
using namespace std;
using namespace std::tr1;
#define opt ios_base::sync_with_stdio(0)
#define lli long long int
#define ulli unsigned long long int
#define I int
#define S string
#define D double
#define rep(i,a,b) for(i=a;i<b;i++)
#define repr(i,a,b) for(i=a;i>b;i--)
#define in(n) scanf("%lld",&n)
#define in2(a,b) scanf("%lld %lld",&a,&b)
#define in3(a,b,c) scanf("%lld %lld %lld",&a,&b,&c)
#define out(n) printf("%lld\n",n)
#define inu(a) scanf("%lld",&a)
#define outu(a) printf("%llu",a)
#define ins(s) scanf("%s",&s)
#define outs(s) printf("%s",s)
#define mod 1000000007
#define inf 100000000000000
typedef long long       ll;
typedef pair<lli,lli>  plli;
typedef vector<lli>     vlli;
typedef vector<ulli>    vulli;
typedef vector<ll>      vll;
typedef vector<string>  vs;
typedef vector<plli>     vplli;
#define MM(a,x)  memset(a,x,sizeof(a));
#define ALL(x)   (x).begin(), (x).end()
#define P(x)       cerr<<"{"#x<<" = "<<(x)<<"}"<<endl;
#define P2(x,y)       cerr<<"{"#x" = "<<(x)<<", "#y" = "<<(y)<<"}"<<endl;
#define P3(x,y,z)  cerr<<"{"#x" = "<<(x)<<", "#y" = "<<(y)<<", "#z" = "<<(z)<<"}"<<endl;
#define P4(x,y,z,w)cerr<<"{"#x" = "<<(x)<<", "#y" = "<<(y)<<", "#z" = "<<(z)<<", "#w" = "<<(w)<<"}"<<endl;
#define PP(x,i)     cerr<<"{"#x"["<<i<<"] = "<<x[i]<<"}"<<endl;
#define TM(a,b)     cerr<<"{"#a" -> "#b": "<<1000*(b-a)/CLOCKS_PER_SEC<<"ms}\n";
#define UN(v)    sort(ALL(v)), v.resize(unique(ALL(v))-v.begin())
#define mp make_pair
#define pb push_back
#define f first
#define s second
#define sz() size()
#define nl cout<<"\n"
#define MX1 100005
#define MX2 1000005
#define bs binary_search
#define lb lower_bound
#define ub upper_bound
lli dx[]={0,0,-1,1,-1,-1,1,1};
lli dy[]={1,-1,0,0,1,-1,-1,1};
lli power(lli a,lli b)
    {
    lli value;
    if(b==0)
        {
        return 1;
    }
    else if(b%2==0)
        {
        value=power(a,b/2)%mod;
        return(value*value)%mod;
    }
    else
        {
        value=power(a,b/2)%mod;
        return ((a*value)%mod*(value))%mod;
    }
}
int main()
{
      /*  #ifndef ONLINE_JUDGE
        freopen("input.txt","r",stdin);
        freopen("output.txt","w",stdout);
        #endif*/
    opt;
    lli n,a,b,ans=0,i,k;
    cin>>n>>k;
    /*if(k==0)
    {
        cout<<n*n;
        exit(0);
    }*/
    rep(i,0,1+n)
    {
        if(i<=k)
        {

        }
        else
        {
            ans+=(i-k)*(n/i);
            lli temp=n%i;
            ans+=(k)?temp-k+1:temp;
//            cout<<i<<" "<<(i-k)*(n/i)+max(0LL,temp-k)<<endl;
        }
    }
    cout<<ans;
}