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)


// Problem : C - A x B + C
// Contest : AtCoder - AtCoder Beginner Contest 179
// URL : https://atcoder.jp/contests/abc179/tasks/abc179_c
// Memory Limit : 1024 MB
// Time Limit : 2000 ms
// Powered by CP Editor (https://github.com/cpeditor/cpeditor)

#include <bits/stdc++.h>
using namespace std;
#define ll long long int
#define endl '\n'
#define pb push_back
#define mod 1000000007
ll power(ll x, ll y) 
{ 
    if (y == 0) 
        return 1; 
    ll p = power(x, y/2) % mod; 
    p = (p * p) % mod; 
  
    return (y%2 == 0)? p : (x * p) % mod; 
} 
ll factorial[1000007];
void fact(ll n)
{
    memset(factorial,0,sizeof(factorial));
    factorial[0]=1;
factorial[1]=1;

    for(int i=1;i<n;i++)
    {
        factorial[i]=(factorial[i-1]*i)%mod;
    }
    
}
ll ncr(ll n,ll r)
{
    ll ans=1;
    ll i;
    ans=factorial[n];
    ll ans1=power(factorial[n-r],mod-2);
    ll ans2=power(factorial[r],mod-2);
    ans=((ans*ans1)%mod*ans2)%mod;
    return ans;
}
ll gcd(ll a, ll b) 
{ 
    if (b == 0) 
        return a; 
    return gcd(b, a % b);  
      
} 

bool prime[1000006];
ll spf[1000006];
void sieve(ll n){
    memset(prime,true,sizeof(prime));
    memset(spf,0,sizeof(spf));
	prime[1]=false; 
    for (ll p=2;p*p<=n;p++){
        if (prime[p]){  
            for (ll i=p*p;i<=n;i+=p){
                prime[i]=false;
                if(spf[i]==0)
                {
                	spf[i]=p;
                }
                }
                
        }
    }
}
void bfs(vector<int> v[],int o,int n,int dp[]){
	queue<int> q;
	q.push(o);
	dp[o]=0;
	bool visited[n+1]={0};
	visited[o]=true;
	while(q.empty()==false){
		int a=q.front();
		q.pop();
		for(int i=0;i<v[a].size();i++){
			if(visited[v[a][i]]==false){
				visited[v[a][i]]=true;
				q.push(v[a][i]);
				dp[v[a][i]]=dp[a]+1;
			}
		}
	}
}
int nov[1000001];
bool visited[1000001];
int dp[1000001];
int dfs(vector<int> v[],int o)
{
    if(visited[o]==0)
    {
        visited[o]=1;
        for(int i=0;i<v[o].size();i++)
        {
            if(visited[v[o][i]]==0)
            {
            dp[v[o][i]]=dp[o]+1;
            nov[o]+=dfs(v,v[o][i]);
            }
        }
        return nov[o]+1;
    }
    else
    return 0;
}
// string s=bitset<64>(n).to_string();
int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
	ll n;
	cin>>n;
	sieve(n+4);
	for(int i=2;i<=n;i++)
	{
		if(spf[i]==0)
		spf[i]=i;
	}
	spf[1]=2;
	// cout<<spf[2]<<endl;
	ll ans=0;
	for(int i=n-1;i>=1;i--)
	{
		
		ll temp=1;
		int j=i;
		while(j%spf[j]==0)
		{
			int x=spf[j];
			ll cnt=0;
			while(j%x==0)
			{
				cnt++;
				j=j/x;
			}
			temp=temp*(cnt+1);
		}
		ans+=temp;
	}
	cout<<ans<<endl;
	return 0;
}