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)

#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,sse3,sse4,popcnt,abm,mmx")

#include <bits/stdc++.h>   
#include <stdio.h>    
 
using namespace std;
     
#define F first
#define S second
#define lb lower_bound               
#define ub upper_bound
#define pb push_back
#define pf push_front    
#define ppb pop_back
#define mp make_pair
#define bpp __builtin_popcountll                                                                                        
#define sqr(x) ((x) * (x)) 
#define al 0x3F3F3F3F
#define sz(x) (int)x.size()
#define all(x) x.begin(), x.end()
#define in insert
#define ppf pop_front
#define endl '\n'
//#define int long long
 
typedef unsigned long long ull;
typedef long long ll;
typedef long double ld;
typedef pair <int, int> pii;
 
const int mod = (int)1e9 + 7;
const int N = (int)2e5 + 123;
const ll inf = (ll)3e18 + 1;

const double pi = acos (-1.0);
const double eps = 1e-7;
const int dx[] = {0, 0, 1, 0, -1};
const int dy[] = {0, 1, 0, -1, 0};

ll a, b, t;

inline void boost () {                    
   ios_base :: sync_with_stdio (NULL);
   cin.tie (NULL), cout.tie (NULL);          
}                                                    

inline void Solve () {
   cin >> a >> b >> t;
   cout << (t / a) * b;
   
}                                     

int main () {                                       
// freopen (".in", "r", stdin);    
// freopen (".out", "w", stdout);              
   boost ();
   int tt = 1;
// cin >> tt;   
   while (tt --) {
      Solve ();
   }                                               
   return 0;      
}