HackerRank October Challenge 2013 Editorial [Chocolate Feast]

Problem: Chocolate Feast
You have $N, price of each chocolate is $C, and you can get one free chocolate for each M wrapper. print maximum chocolate you can eat.

Solution Idea: first count how many chocolates you can buy with $N. it simply N/C.
eat all chocolates, now you have the wrappers. if you have more than M wrappers, return multiple of M wrappers, let, M*X wrappers where X>0 to get X more free chocolates. eat them, now count how many wrapper you have, continue the procedure until you don't have enough wrapper to have a free chocolate.

Example:  for N=10, C=2, M=3.
you can buy 5 chocolate with $10, now you have 5 wrapper. return 3 wrapper to get a free chocolate, now you have 2 wrapper and 1 wrapper after eating the free chocolate. so total wrapper=3, return 3 wrapper to get another free chocolate.
so, total chocolate= 5 (buy) + 2 (free) = 7.

Code: 

1. #include<bits/stdc++.h>
2. using namespace std;
3.  
4. #define SET(a) memset(a,-1,sizeof(a))
5. #define CLR(a) memset(a,0,sizeof(a))
6. #define PI acos(-1.0)
7.  
8. int main()
9. {
10. int tc, n, c, m, tch, wr, tmp;
11. cin>>tc;
12. while(tc--)
13. {
14. cin>>n>>c>>m;
15. tch=wr=n/c;
16. while(wr>=m)
17. {
18. tmp=wr/m;
19. wr=wr%m;
20. wr+=tmp;
21. tch+=tmp;
22. }
23. cout<<tch<<endl;
24. }
25. return 0;
26. }

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