## Problem: Halloween party

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.

This is my code:

#include <stdio.h>

long _Halloween_Party(long _n);

int main() {

/* Enter your code here. Read input from STDIN. Print output to STDOUT */

/*n is the amount of test cases*/

int _n;/*array a to store input value, b to store result to print*/

long _a[1000],_b[1000],_i;

scanf(“%d”,&_n);

for(_i=0;_i<_n;++_i)

scanf(“%ld”,&_a[_i]);

for(_i=0;_i<_n;++_i){

_b[_i]=_Halloween_Party(_a[_i]);

}

for(_i=0;_i<_n;++_i){

printf(“%ld”,_b[_i]);

if(_i!=_n-1)

printf(“\n”);

}

return 0;

}/*Write down some case in paper, and you’ll find out your own way to solve this problem*/

/*This is my method*/

long _Halloween_Party(long _n){

long _i,_Count=0;

long _Sum=0,_Raise=1;

for(_i=1;_i<_n;++_i){

if(_Count==2){

_Raise+=1;

_Count=0;

}

_Sum+=_Raise;

_Count+=1;

}

return _Sum;

}

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