Problem:
Score : 300 points

Problem Statement

Given is a positive integer N.
How many tuples (A,B,C) of positive integers satisfy A \times B + C = N?

Constraints

2 \leq N \leq 10^6

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N

Output

Print the answer.

Sample Input 1

3

Sample Output 1

3

There are 3 tuples of integers that satisfy A \times B + C = 3: (A, B, C) = (1, 1, 2), (1, 2, 1), (2, 1, 1).

Sample Input 2

100

Sample Output 2

473

Sample Input 3

1000000

Sample Output 3

13969985

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define LI list<int>
#define vct vector
#define D(a) (double)(a)
#define vi vct<int>
#define pii pair<int,int>
#define pll pair<ll,ll>
#define vvi vct<vi>
#define vll vct<ll>
#define vvll vct<vll>
#define vpii vct<pii>
#define vpll vct<pll>
#define vb vct<bool>
#define vs vector<string>
#define all(a) a.begin(),a.end()
#define allr(a) a.rbegin(),a.rend()
#define mp(a,b) make_pair(a,b)
#define pb push_back
#define ff first
#define ss second
#define bg begin()
#define UNIQUE(X) (X).erase(unique(all(X)),(X).end())
#define ft cout<<"for test"<<endl;
#define read(v,a,n) for(int i=a; i<n; i++)cin>>v[i];
#define print(v) for(auto it:v)cout<<it<<' ';cout<<endl;
#define PI 2*acos(0.0)
#define yes cout<<"YES"<<endl
#define no cout<<"NO"<<endl
#define FIO ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define t_c int test,cs=1;cin>>test;while(test--)
///................function.....................///
#define siz(s) (int)(s.size())
#define min3(a,b,c) min(a,min(b,c))
#define max3(a,b,c) max(a,max(b,c))
#define mod 1000000007
int main()
{
    FIO;
    ///freopen("in.txt","r",stdin);
    ///freopen("output.txt","w",stdout);
    vi pre(1000005);
    ll i,j;
    ll nn=1000001;
    for(i=1; i<nn; i++)
    {
        for(j=i; j*i<nn; j++)
        {
            if(i!=j)
                pre[i*j]++;
            pre[i*j]++;
        }
    }
    ll n;
    cin>>n;
    ll ans=0;
    for(i=1; i<n; i++)
    {
        ll tmp = pre[i];
ans+=tmp;
    }
    cout<<ans<<endl;
}