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 FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n) FOR(i,0,n)
#define ALL(v) begin(v),end(v)
#define fi first
#define se second
template<typename A, typename B> inline bool chmax(A &a, B b) { if (a<b) { a=b; return 1; } return 0; }
template<typename A, typename B> inline bool chmin(A &a, B b) { if (a>b) { a=b; return 1; } return 0; }
using ll = long long;
using pii = pair<int, int>;
constexpr ll INF = 1ll<<30;
constexpr ll longINF = 1ll<<60;
constexpr ll MOD = 1000000007;
constexpr bool debug = 0;
//---------------------------------//

vector<ll> find_divisor(ll n) {
	vector<ll> divisor;
	
	ll i = 1;
	for (; i * i <= n; i++) if (n % i == 0) divisor.push_back(i);
	i--;
	if (i * i == n) i--;
	for (; i >= 1; i--) if (n % i == 0) divisor.push_back(n / i);
	
	return divisor;
}

int main() {
	int N;
	cin >> N;
	
	ll ans = 0;
	
	vector<int> cnt(N + 1);
	FOR(i, 1, N) FOR(j, 1, N) {
		if (i * j > N) break;
		++cnt[i * j];
	}
	
	FOR(c, 1, N) {
		int diff = N - c;
		ans += cnt[diff];
	}
	cout << ans << endl;
	return 0;
}