## Problem: Chocolate Feast

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.

Code:

#include <stdio.h>

long Chocolate_Feast(long a,long b,long c);

int main() {

/* Enter your code here. Read input from STDIN. Print output to STDOUT */
int n,i;
long N,C,M,a[1001];
scanf(“%d”,&n);
for(i=0;i<n;++i){
scanf(“%ld%ld%ld”,&N,&C,&M);
a[i]=Chocolate_Feast(N,C,M);
}
/*As usual, this fragment will help us avoid leaving a redundant character ‘\n’*/
for(i=0;i<n;++i){
printf(“%ld”,a[i]);
if(i!=n-1)
printf(“\n”);
}
return 0;
}

long Chocolate_Feast(long a,long b,long c){
long Amount,Wrapper,Temp;
Amount=a/b;
Wrapper=Amount;
/*The condition to eliminate the while loop is when the wrapper less than c*
/*c is the amount of wrappers that used to exchange a new chocolate*/
while((Wrapper/c)>0){
Temp=Wrapper/c;
/*Compute the number of new wrappers, always less than the previous step*/
Wrapper=Wrapper%c + Temp;
Amount+=Temp;
}
return Amount;
}