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 <iostream>
#include <cstdint>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <vector>
#include <list>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <cctype>
#include <cassert>
#include <climits>
#include <string>
#include <bitset>
#include <cfloat>
#include <unordered_set>
#pragma GCC optimize("Ofast")
using namespace std;
typedef long double ld;
typedef long long int ll;
typedef unsigned long long int ull;
typedef vector<int> vi;
typedef vector<char> vc;
typedef vector<bool> vb;
typedef vector<double> vd;
typedef vector<string> vs;
typedef vector<ll> vll;
typedef vector<pair<int, int>> vpii;
typedef vector<vector<int>> vvi;
typedef vector<vector<char>> vvc;
typedef vector<vector<string>> vvs;
typedef vector<vector<ll>> vvll;

#define rep(i, n) for (ll i = 0; i < (n); ++i)
#define rrep(i, n) for (ll i = 1; i <= (n); ++i)
#define irep(it, stl) for (auto it = stl.begin(); it != stl.end(); it++)
#define drep(i, n) for (ll i = (n)-1; i >= 0; --i)
#define mes(a) cout << (a) << endl
#define dmes(a, b) cout << (a) << " " << (b) << endl
#define re0 return 0
#define mp(p, q) make_pair(p, q)
#define pb(n) push_back(n)
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
#define Sort(a) sort(a.begin(), a.end())
#define rSort(a) sort(a.rbegin(), a.rend())
#define Rev(a) reverse(a.begin(), a.end())
#define popcnt(n) __builtin_popcount(n)
#define MATHPI acos(-1)
#define itn int
int dx[8] = {1, 0, -1, 0, 1, -1, -1, 1};
int dy[8] = {0, 1, 0, -1, 1, 1, -1, -1};
template <class T>
inline bool chmax(T &a, T b)
{
    if (a < b)
    {
        a = b;
        return 1;
    }
    return 0;
}
template <class T>
inline bool chmin(T &a, T b)
{
    if (a > b)
    {
        a = b;
        return 1;
    }
    return 0;
}
struct io
{
    io()
    {
        ios::sync_with_stdio(false);
        cin.tie(0);
    }
};
const int INF = INT_MAX;
const ll LLINF = 1LL << 60;
const ll MOD = 1000000007;
const double EPS = 1e-19;


const int NMAX=1000001;


signed main() {
    ll n;
    cin >> n;
    
    ll r = 0;
    
    rrep(i, 1e6) {
        rrep(j, 1e6/i) {
            if (i * j < n) r++;
        }
    }
    
    mes(r);
 
    
}