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>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/detail/standard_policies.hpp>

using namespace std;
using namespace __gnu_pbds;
long long power(long long x, long long y, long long mood) {
	if (y == 0)
		return 1;
	long long ret = power(x, y >> 1, mood) % mood;
	if (y % 2 == 0)
		return (ret * ret) % mood;
	else
		return (x * (ret * ret % mood)) % mood;
}
const double Eps = (1e-9);
int dcmp(double x, double y) {
	return fabs(x - y) <= Eps ? 0 : x < y ? -1 : 1;
}
typedef tree<int, null_type, less<int>, rb_tree_tag,
		tree_order_statistics_node_update> ordered_set;
typedef tree<pair<int, int>, null_type, less<pair<int, int>>, rb_tree_tag,
		tree_order_statistics_node_update> ordered_setpair;
typedef long long ll;
typedef vector<int> vi;
typedef vector<ll> vll;
typedef vector<string> vs;
typedef vector<pair<int, int> > vii;
typedef map<int, int> mii;
typedef map<string, int> msi;
typedef pair<int, int> pii;
#define mii map <int, int>
#define mll map <ll, ll>
#define pii pair <int, int>
#define pll pair <ll, ll>
#define vi vector <int>
#define vd vector <double>
#define vll vector <ll>
#define sll set <ll>
#define spii set <pii>
#define vpii vector <pair <int,int> >
#define vpll vector <pair <long long,long long> >
#define vvi vector <vector <int> >
#define vvpii vector <vector <pii > >
#define sfi1(v) scanf("%d",&v)
#define sfi2(v1,v2) scanf("%d %d",&v1,&v2)
#define sfi3(v1,v2,v3) scanf("%d %d %d",&v1,&v2,&v3)
#define sfll1(v) scanf("%I64d",&v);
#define sfll2(v1,v2) scanf("%I64d %I64d",&v1,&v2)
#define sfll3(v1,v2,v3) scanf("%I64d %I64d %I64d",&v1,&v2,&v3)
#define min3(a,b,c)     min(a,min(b,c))
#define max3(a,b,c)     max(a,max(b,c))
#define min4(a,b,c,d)   min(a,min(b,min(c,d)))
#define max4(a,b,c,d)   max(a,max(b,max(c,d)))
#define el	'\n'
#define lp(i,n) for(int i = 0;i < int(n);i++)
#define lp1(i,n) for(int i = 1;i <= int(n);i++)
#define rlp(i, n) for (int i = (n) - 1; i >= 0; i--)
#define rng(i,a,b) for(int i = int(a);i <= int(b);i++)
#define mp(x,y)  	make_pair((x),(y))
#define pb(x)    	push_back((x))
#define pob()    	pop_back()
#define all(v)  	((v).begin()),((v).end())
#define sz(x)  		((int) (x).size())
#define bits(a) 	__builtin_popcountll(a)
#define fr first
#define sc second
#define pii pair<int, int>
#define sortva(v) sort(all(v))
#define sortvd(v) sort(v.rbegin(),v.rend())
#define sortaa(a,n) sort(a,a+n)
#define sortad(a,n) sort(a,a+n),reverse(a,a+n)
#define SS stringstream
#define ll long long
#define clr(x, y) memset(x, y, sizeof(x))
#define foreach(it,x) for(__typeof__((x).begin()) it = (x).begin(); it != (x).end(); ++it)
#define read freopen("input.in", "rt", stdin);
#define write freopen("output.in", "wt", stdout);
#define fastIO cout << fixed << setprecision(7), ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);
const long long key1 = (31);
const long long key2 = 59;
int dx[] = { 0, 1, 0, -1, 1, -1, 1, -1 };
int dy[] = { 1, 0, -1, 0, -1, 1, 1, -1 };
int kx[] = { 2, 2, 1, -1, -2, -2, 1, -1 };
int ky[] = { 1, -1, 2, 2, 1, -1, -2, -2 };
int gcd(int a, int b) {
	if (a == 0)
		return b;
	return gcd(b % a, a);
}
ll lcm(ll a, ll b) {
	return (a / gcd(a, b)) * b;
}
int setBit(int num, int idx, int val) {
	return val ? (num | (1 << idx)) : (num & (~(1 << idx)));
}
int getBit(int num, int idx) {
	return ((num >> idx) & 1) == 1;
}
int const N = 5e6+5 , mod = 100000000 , oo = 1e9 + 10;
//-------------------------------------------------------------------------------------

int cnt[N];
int main()
{

	int n;
	cin>>n;
	for (int i=1;i<=1e6;i++){
		for (int j=i;j<=1e6;j+=i){
			cnt[j]++;
		}
	}
	ll sum=0;

	lp1(i,n){
	sum+=cnt[n-i];
	}
	cout<<sum;













/*
*/
return 0;
}