C++

Chocolate Feast

PROBLEM

Little Bob loves chocolates and goes to the store with a $N bill with $C being the price of each chocolate. In addition, the store offers a discount: for every M wrappers he gives the store, he’ll get one chocolate for free. How many chocolates does Bob get to eat?

Input Format:

The first line contains the number of test cases T (<=1000).
Each of the next T lines contains three integers N, C and M


Output Format:

Print the total number of chocolates Bob eats.

Constraints:

2 <= N <= 100000
1 <= C <= N
2 <= M <= N

Sample input

3
10 2 5
12 4 4
6 2 2

Sample Output

6
3
5




Explanation 

In the first case, he can buy 5 chocolates with $10 and exchange the 5 wrappers to get one more chocolate thus making the total number of chocolates he can eat as 6

In the second case, he can buy 3 chocolates for $12. However, it takes 4 wrappers to get one more chocolate. He can’t avail the offer and hence the total number of chocolates remains 3.

In the third case, he can buy 3 chocolates for $6. Now he can give 2 of this 3 wrappers and get 1 chocolate. Again, he can use his 1 unused wrapper and 1 wrapper of new chocolate to get one more chocolate. Total is 5.

PROGRAM                                              OUTPUT

#include <iostream>
using namespace std;
int main() {
  int t,n,c,m,rc,nc,fc,i,*a,nvc;
  cin>>t;
  a=new int[t];
  for(i=0;i<t;i++)
  {
     n=c=m=0;
     cin>>n>>c>>m;
     fc=nc=rc=nvc=0;
     rc=nc=(n/c);
     while(rc>=m)
     {
	fc=(rc/m);
        nvc=nvc+fc;
        rc=(rc%m)+fc;
     }
     a[i]=nc+nvc;
  }
  for(i=0;i<t;i++)
  {
     cout<<a[i]<<endl;
  }
	 
  return 0;
}
3


10 2 5 12 4 4 6 2 2


6 3 5