Problem:
Score : 100 points

Problem Statement

A biscuit making machine produces B biscuits at the following moments: A seconds, 2A seconds, 3A seconds and each subsequent multiple of A seconds after activation.

Find the total number of biscuits produced within T + 0.5 seconds after activation.

Constraints

All values in input are integers.

1 \leq A, B, T \leq 20

Input

Input is given from Standard Input in the following format:

A B T

Output

Print the total number of biscuits produced within T + 0.5 seconds after activation.

Sample Input 1

3 5 7

Sample Output 1

10

Five biscuits will be produced three seconds after activation.

Another five biscuits will be produced six seconds after activation.

Thus, a total of ten biscuits will be produced within 7.5 seconds after activation.

Sample Input 2

3 2 9

Sample Output 2

6

Sample Input 3

20 20 19

Sample Output 3

0

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>

using namespace std;

typedef long long int ll;
typedef long long int lli;
typedef unsigned long long int ull;
typedef long double ld;
typedef string str;

typedef vector<char> vc;
typedef vector<bool> vb;
typedef vector<ld> vd;
typedef vector<str> vs;
typedef vector<ll> vll;
typedef vector<pair<ll, ll>> vpll;
typedef vector<vector<ll>> vvll;
typedef vector<vector<char>> vvc;
typedef vector<vector<str>> vvs;

#define rep(i, n) for (ll i = 0; i < (n); ++i)
#define rrep(i, n) for (ll i = 1; i <= (n); ++i)
#define fin(ans) cout << (ans) << endl;
//３つ以上の値をセットに(3より多いときはカッコ内を増やす)\
sort -> p -> q -> r
#define mt(p, q, r) make_tuple(p, q, r)
#define all(a) a.begin(), a.end()

template <typename T>
void V(T &n)
{
  for (ll i = 0; i < n.size(); i++)
  {
    cin >> n[i];
  }
}

template <typename T, typename U>
void V2(T &n, U &m)
{
  for (ll i = 0; i < n.size(); i++)
  {
    cin >> n[i] >> m[i];
  }
}

template <typename T>
void Sort(T &n, ll a = 0, ll b = 0)
{
  if (b == 0)
  {
    return (sort(n.begin() + a, n.end()));
  }
  else
  {
    return (sort(n.begin() + a, n.begin() + b));
  }
}

template <typename T>
void Down(T &n, ll a = 0, ll b = 0)
{
  if (b == 0)
  {
    return (sort(n.begin() + a, n.end(), greater<ll>()));
  }
  else
  {
    return (sort(n.begin() + a, n.begin() + b, greater<ll>()));
  }
}

void Main()
{
  ll a, b, t; cin >> a >> b >> t;
  fin(t / a * b)
}

int main()
{
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout << fixed << setprecision(20); //高精度少数表示
  Main();
  return 0;
}