- Practice
- Algorithms
- Implementation
- Chocolate Feast
- Discussions

# Chocolate Feast

# Chocolate Feast

mbkr1992 + 24 comments # Only 1 output worng among all cases?

Case No. 5, input string is "43203 60 5", my output is "892", right answer is "899" I just want to make sure whether there is a mistake at your end. Kindly check, I've checked plenty of time, If not Kindly explain the case? Much appreciaited.

dheeraj + 13 comments `43203 / 60 = 720 chocolates ( 3 dollars left and you cannot buy any more chocolate for 3 dollar ) 720 / 5 = 144 chocolates ( we can buy 144 chocolates from 720 wrappers ) 144 / 5 = 28 chocolates ( we can buy 28 chocolates from 144 wrappers, 4 wrappers left ) (28 + 4 ) / 5 = 6 ( 28 wrappers + 4 wrappers left from 144, we can buy 6 chocolates ) (6 + 2) / 5 = 1 ( 6 wrappers + 2 wrappers left from 32 wrappers, we can buy 1 chocolate ) This sums to 720 + 144 + 28 + 6 + 1 = 899.`

SudeepSrivastav + 0 comments [deleted]pulkitcoder + 0 comments Same problem with my code also.

alejandro33 + 0 comments thaks I miss that case

Storm__Shadow + 0 comments [deleted]Storm__Shadow + 0 comments [deleted]Storm__Shadow + 0 comments [deleted]Storm__Shadow + 0 comments [deleted]koushik_077 + 0 comments check out this code .. https://www.hackerrank.com/challenges/chocolate-feast/submissions/code/20066920

abhishekramkumar + 1 comment this kinda worked import java.io.

*; import java.util.*; import java.text.*; import java.math.*; import java.util.regex.*;public class Solution {

`public static void main(String[] args) { Scanner in = new Scanner(System.in); int t = in.nextInt(); for(int a0 = 0; a0 < t; a0++){ int n = in.nextInt(); int c = in.nextInt(); int m = in.nextInt(); int totalChoc = n / c; int totalWrappers = totalChoc; int freeChocs = 0; while (m <= totalWrappers) { int wrapperTogive=totalWrappers-(totalWrappers%m) ; freeChocs = totalWrappers / m; totalWrappers = (totalWrappers - wrapperTogive) + freeChocs; totalChoc=totalChoc + freeChocs; } System.out.println(totalChoc); } }`

}

shashwatkhanna31 + 3 comments I guess you are doing way to much unnecessary work. Heree is my while loop

while(m<=wrappers){ wrappers = wrappers - m; totalchocs++; wrappers++; }

jason_e_reynolds + 4 comments This approach is clean and easy to follow, which is great.

However, it results in O(N) time (input of 10000 1 2 results in the loop running 9999 times.)

If you switch the loop to the following you end up with O( lg(N)) (or 14 in the worst case, since you are using integer math).

def chocolateFeast(cash, cost, wrapper_cost): chocolate = cash//cost wrappers = chocolate while wrappers // wrapper_cost > 0: new_chocolate = wrappers // wrapper_cost wrappers -= new_chocolate * wrapper_cost wrappers += new_chocolate chocolate += new_chocolate return chocolate

akonibrahim1 + 0 comments Hi Jason I kind of did the same thing what you did but mine didn't pass the test cases how whats the difference

import time

start_time = time.time()

def chocolateFeast1(he_has,cost,wrapper):

`current_chocates = int(he_has / cost) current_wrappers = current_chocates while current_wrappers >= wrapper: current_chocates += 1 current_wrappers -= cost current_wrappers += 1 print(f"Took { time.time() - start_time}") print (current_chocates)`

start_time = time.time() def chocolateFeast(cash, cost, wrapper_cost):

`chocolate = cash//cost wrappers = chocolate while wrappers // wrapper_cost > 0: new_chocolate = wrappers // wrapper_cost wrappers -= new_chocolate * wrapper_cost wrappers += new_chocolate chocolate += new_chocolate print(f"Took { time.time() - start_time}") print (chocolate)`

rampappula + 2 comments wrappers = wrappers % wrapper_cost instead of wrappers -= new_chocolate * wrapper_cost

was better understandable for me.

pcortezzi + 0 comments Your code is flawless. Quick, simple and completely readable! Congratulations!

frankdeadlycaste + 0 comments # see this

def chocolateFeast(n, c, m):

`ch = n/c all_ch = ch while (ch>=1): all_ch += ch/m ch = ch/m + ch%m return all_ch`

manikondapraveen + 0 comments your code will result in performance issues when the count is too high. Thats the reason suggested to do with / and % operators

agwolgemuth + 1 comment I did something similar. As several others have noted, it's an inefficient approach, but it was really quick to write and handled all the test cases so I considered it fit for purpose.

firozbhaikardar1 + 0 comments int chocolateFeast(int n, int c, int m) { int i,sum,x,j; sum=n/c; x=sum; if(sum==0) return 0; while(x>=m) { j=x%m; i=x/m; sum=sum+i; x=i+j; } return sum; }

alokmhn5 + 0 comments there is problem with ur code. 899 is correct.

dropyghost + 0 comments any idea if 3 dollars left are carry to next trip?

iainws + 0 comments thanks. I had a terrible time with this problem. maybe it was the noise in my environment, maybe it was something to do with the computer i'm using... I couldn't get it exactly right untill I read this. Thanks for the code.

hrshtvrm + 1 comment Why does this time out?

import java.util.*; public class q{ public static void main(String args[]){ Scanner sc=new Scanner(System.in); int t=sc.nextInt(); for(int g=0;g<t;g++){ int n=sc.nextInt(),c=sc.nextInt(),m=sc.nextInt(); int totalChoc=0;int wrappers=n/c; while(wrappers>0){ totalChoc=n/c+wrappers/m; wrappers=wrappers%m;} System.out.println(totalChoc);}}}

abishek_surpur2 + 3 comments `def chocolateFeast(n, c, m): ch=n//c wr=ch while(wr>=m): ch+=wr//m wr=wr//m+wr%m return ch`

try this out with time comlexity less than O(n) :)

shivam_hackgod + 1 comment can you explain why my code is not giving output for testcase 8 only.

def chocolateFeast(n, c, m): inichoc=n//c total=inichoc leftwrap=0 i=0 while inichoc+leftwrap>=m: addchoc=math.floor((inichoc+leftwrap)/m) leftwrap=(inichoc-i)%m total+=addchoc inichoc=addchoc i+=1 return total

mike_buttery + 0 comments [deleted]

petia_davidova + 0 comments What is the complexity of this? I have the same solution but I am not sure what my complexity is, I find it a bit difficult. :)

frankdeadlycaste + 1 comment # try this

def chocolateFeast(n, c, m):

`ch = n/c all_ch = ch while (ch>=1): all_ch += ch/m ch = ch/m + ch%m return all_ch`

Gautham_B_A + 1 comment There is no doubt about the veracity of the answers. Do not falter. Try again. I too faced the same issue. Try to code it exactly as the problem has been described.

pankaj113 + 2 comments I tried again :-)

import java.io.

*; import java.util.*; import java.text.*; import java.math.*; import java.util.regex.*;public class Solution {

`public static void main(String[] args) { Scanner in = new Scanner(System.in); int t = in.nextInt(); for(int a0 = 0; a0 < t; a0++){ int n = in.nextInt(); int c = in.nextInt(); int m = in.nextInt(); int chocs = n/c; //chocolates bought with Money int wrapper = chocs; //wrappers from chocolate int ex_chocs = wrapper/m; //chocolates from wrappers while(wrapper>=m){ ex_chocs = ex_chocs+(wrapper/m + wrapper%m)/m; wrapper = wrapper/m; } System.out.println(chocs+ex_chocs); } }`

}

cesarjom + 3 comments @pankaj113, You can simplify/reduce your code by moving this line

`int ex_chocs = wrapper / m;`

inside while loop, like so:

`while(wrapper >= m) { int ex_chocs = wrapper / m; wrapper = ex_chocs + wrapper % m; chocs += ex_chocs; } System.out.println(chocs);`

avi_ganguly94 + 0 comments just th tweak i needed

narayan_m + 0 comments can be further simplified

while(w>=m){ chocs+=w/m; w=w/m+(w%m); }

koushik_077 + 0 comments

enprush + 1 comment Against my better judgement, wanting a one line answer I got stuck here :D Hahahah

wittrup + 3 comments Yeah, I cannot see a way to make this a one-liner, can you?

for _ in range(int(input())): n,c,m=map(int,input().split()) print(n//c+(n//c-1)//(m-1))

bitSurfer151093 + 0 comments Dude! This was amazing. Do you think you can break it to be able to understand?

beckwithdylan + 0 comments [deleted]beckwithdylan + 1 comment Here's your one-liner

print(*(n//c+(n//c-1)//(m-1) for n,c,m in [map(int,input().split()) for _ in range(int(input()))]),sep='\n')

vlomako + 1 comment It is the same solution. And it is 100 times less readable for the sake of two lines of code.

P.S. Most of languages, other than Python allow converting programs into one line simply by deleting the newline characters. But nobody does that.

gwarks + 0 comments Python also allows that

`for _ in range(int(input())):n,c,m=map(int,input().split());print(n//c+(n//c-1)//(m-1))`

But for some people is no more oneline because it cant used in another line and produce no return value

hyadav2606 + 0 comments check my link for solution..its in python http://ideone.com/Al2fEX

great_coder + 4 comments Following code works for me:

wrappers = n/c;

count = wrappers;

while (wrappers >= m) {

wrappers = wrappers - m;

count = count + 1;

wrappers = wrappers + 1;

}

System.out.println (count);

Debu_Basu + 1 comment I couldn't figure out the algo. Pls help.

jbonedev + 1 comment First, your total (initial) number of chocolates, and wrappers, is equal to what you can buy with cash, which is n/c. Now you have to iteratively give your wrappers to the shop keeper, in exchange for chocolates, costing you m wrappers (wrappers / m). During this iterative process you add the chocolates to your total count. You subtract the number of wrappers you have handed in and also add back the number of wrappers you receive from the chocolates you are getting back. Here's a java solution.

`public static void main(String[] args) { Scanner in = new Scanner(System.in); int t = in.nextInt(); for(int a0 = 0; a0 < t; a0++){ int n = in.nextInt(); int c = in.nextInt(); int m = in.nextInt(); int total = n/c; int wrappers = total; while(wrappers >= m) { total += (wrappers / m); int leftover = wrappers % m; wrappers = (wrappers / m) + leftover; } System.out.println(total); } }`

TechJack4 + 0 comments That's exactly what I did.

cferguson + 1 comment That's basically what I did, except in Python.

`choc, wrappers = [int(n/c), int(n/c)] while (wrappers >= m): choc+= 1 wrappers -= m wrappers += 1 print(choc)`

SebastianNielsen + 1 comment A better version:

choc = wrappers = n // c while (wrappers >= m): choc+= 1 wrappers -= m - 1 print(choc)

mike_buttery + 0 comments Similar here without iterating every bar/wrapper swap

bars = wrappers = n // c while wrappers >= m: bars += wrappers // m wrappers -= wrappers // m * (m - 1) return bars

Purist238 + 0 comments Same as mine!

renyo + 2 comments i dont get it, why sum plus one on wrappers at the end of the loop... ? that was my mistake...

Deepeka + 0 comments every time you subtract m wrappers you get one chocolate so one more wrapper so you add 1

shankarpenore + 0 comments You are just right

**brother**

sattujaiswal + 4 comments static int cc; public static void main(String[] args) {

`Scanner in = new Scanner(System.in); int t,n,c,m,r; t = in.nextInt(); while(t-->0){ n = in.nextInt(); c = in.nextInt(); m = in.nextInt(); r=n/c; cc=r; while(r>=m){ cc=cc+r/m; r=r%m+r/m; } System.out.println(cc); } }`

}

evilcat + 0 comments I am (currently) just on hackerranks to have "exercises" to learn java/programming with and look at neat solutions and tips after getting the solution my own way and see if i can improve it.

I loved the neatness of the above solution and thought wow that must run so much faster than my solution, so i submitted them one after another a few times only to see my solution outperform the above. Why? Is the remainder operator in the above solution really so expensive and my jumble of code faster just because it doesn't use it?

import java.io.

*; import java.util.*;public class Solution {

`public static void main(String[] args) { Scanner sc = new Scanner(System.in); int t = Integer.parseInt(sc.nextLine()); //t = number of test cases int[][] tc = readInput(sc, t); //tc[t][0] = money. tc[t][1] = price. tc[t][2] = wrappers per free bar for (int i = 0; i<t; i++){ //loop for all test cases int choc = calcChoc(tc,i); //work out how much choc can be bought System.out.println(choc); //print result for the test case } } //calculate how much choc he can buy with m $ at p price with w wrappers needed for a free bar public static int calcChoc(int[][] tc,int i){ int m = tc[i][0]; //money he has int p = tc[i][1]; //price of choc int w = tc[i][2]; //wrappers per free bar int bars = m/p; //how many bars he can buy initially int wrappers = bars; //each bar is a wrapper from initial purpose //loop to turn in all wrappers while it is possible to do so while (w<=wrappers){ int barsFromTurnIn = wrappers/w; //bars from turning in current wrappers. bars = bars + barsFromTurnIn; //new bar count wrappers = wrappers - (barsFromTurnIn * (w-1)); //wrapper count reduced by amount of wrappers turned in -1 wrapper per bar recieved from turn in. if (w==1){ //break out of infinite loop when you get 1 bar for 1 wrapper! System.out.print("Infinite Bars, exiting infinite loop at bars = "); break; } } return bars; } //read input for each test case and make 2d array of the info public static int[][] readInput(Scanner sc, int t){ int[][] input = new int[t][3]; for (int i = 0; i<t; i++){ String[] inputLine = sc.nextLine().split(" "); input[i][0] = Integer.parseInt(inputLine[0]); input[i][1] = Integer.parseInt(inputLine[1]); input[i][2] = Integer.parseInt(inputLine[2]); } return input; }`

}

xinzhg + 0 comments Could you explain your code?

jayparekh56 + 0 comments +1 Used the exact same logic with c++

jayparekh56 + 0 comments +1 Used the exact same logic with c++

Agam_Griffin + 1 comment For checking the visualisation of your code... TRY...

pythontutor.com

You can check the test cases and exact bug in your code. :-)

evilcat + 0 comments [deleted]

SergioGM + 0 comments It works for me:

`public static void main(String[] args) { Scanner in = new Scanner(System.in); int t = in.nextInt(); for(int a0 = 0; a0 < t; a0++){ int n = in.nextInt(); int c = in.nextInt(); int m = in.nextInt(); int s = n/c; int wrapper = n/c; while(wrapper-m>=0){ wrapper-=(m-1); s++; } System.out.println(s); } }`

The trick resides in while you are having a free chocolate, you have another more changeable wrapper. So, while you have wrappers, if you can change it for another choco (wrapper-m>=0), decrease your current wrappers as much as you can deliver for a new choco minus 1. That minus 1 means that you will have another more wrapper coming from your new free choco =)

snmill + 0 comments Why not ?

`static int numberOfChocolates(int money, int price, int bonus) { int chocolades = money / price; MainAndRest accumulated = accumulate(chocolades, bonus); if (accumulated.rest >= bonus) { MainAndRest accumulatedRests = accumulate(accumulated.rest, bonus); int result = chocolades + accumulated.main + accumulatedRests.main; return result; } else { int result = chocolades + accumulated.main; return result; } } static class MainAndRest { int main; int rest; public MainAndRest(int main, int rest) { this.main = main; this.rest = rest; } } static MainAndRest accumulate(int subMoney, int bonus) { int main = subMoney / bonus; int rest = subMoney % bonus; if ((main + rest) < bonus) { return new MainAndRest(main, rest); } else { MainAndRest sub = accumulate(main, bonus); return new MainAndRest(main + sub.main, rest + sub.rest); } }`

snmill + 0 comments Why not ?

`static int numberOfChocolates(int money, int price, int bonus) { int chocolades = money / price; MainAndRest accumulated = accumulate(chocolades, bonus); if (accumulated.rest >= bonus) { MainAndRest accumulatedRests = accumulate(accumulated.rest, bonus); int result = chocolades + accumulated.main + accumulatedRests.main; return result; } else { int result = chocolades + accumulated.main; return result; } } static class MainAndRest { int main; int rest; public MainAndRest(int main, int rest) { this.main = main; this.rest = rest; } } static MainAndRest accumulate(int subMoney, int bonus) { int main = subMoney / bonus; int rest = subMoney % bonus; if ((main + rest) < bonus) { return new MainAndRest(main, rest); } else { MainAndRest sub = accumulate(main, bonus); return new MainAndRest(main + sub.main, rest + sub.rest); } }`

gajeshbhat + 0 comments This Code Passes all the test cases and quite clear with variable names. https://www.hackerrank.com/challenges/chocolate-feast/submissions/code/21403363

mchisty + 0 comments This solution works:

`void main() { int numChocs = n / c; if (numChocs >= m) { numChocs = numChocs + test(numChocs, m); } System.out.println(numChocs); } static int test(int wraps, final int m) { int remainder = wraps % m; int numChocs = wraps / m; if (numChocs + remainder >= m) { numChocs = numChocs + test(numChocs + remainder, m); } return numChocs; }`

Swapnil_008 + 1 comment # include

# include

# include

# include

# include

# include

# include

int main() { int t,i; scanf("%d",&t); for(i=1;i<=t;i++) { int n,c,m,ans,x,sum=0; scanf("%d %d %d",&n,&c,&m); ans=n/c; x=ans; while(x>=m) { sum=sum+(x/m); x=(x/m)+(x%m);

`} printf("%d\n",ans+sum); } return 0;`

}

Astrodynamic + 0 comments #include <stdio.h> int main() { int t; scanf("%d", &t); for (int i = 0; i < t; i++) { int n, c, m; scanf("%d %d %d", &n, &c, &m); int ans, x; ans = n / c; x = ans; int sum = 0; while (x >= m) { sum += (x / m); x = (x / m) + (x % m); } printf("%d\n", ans + sum); } return 0; }

trigodeepak + 0 comments we need to decrement number of wrappers as we use them

raj_yadav29oct + 0 comments try this -

int wrap = m; bool flag = true; int choc = 0; while(wrap>=m) { if(flag) { wrap = 0; flag = false; choc += n / c; wrap = n / c; } else { choc += wrap / m; if (wrap == m) wrap = 0; else wrap = (wrap / m) + (wrap % m); } } Console.WriteLine(choc);

hirenmangukiya86 + 2 comments simple solution in c++

int main(){ int t; cin >> t; for(int a0 = 0; a0 < t; a0++){ int n; int c; int m; int t=0,w=0,l=0; cin >> n >> c >> m; t=n/c; w=n/c; while(w>=m) { t+=w/m; w=(w/m) + (w%m); } cout<<t<<endl; } return 0; }

danyfamily22 + 0 comments thanks broo

agamyaji3 + 1 comment can you please explain how this works ?

vlomako + 0 comments So, how to solve the problem in two operators?

Bobby uses money on the first iteration only. So, after 'n //= c' operator there are only two numbers in use: n - number of bars/wrappers Bobby has at the beginning, m - price of the promotional bar in wrappers.

We can get rid of iterations when understanding that the chockolate bar "costs" not m wrappers, but m-1, because Bobby has one additional wrapper for every m of them. So the promotional bars number can be calculated without 'for/while' operator, simply dividing wrapper number by m-1.

But Bobby can't spend all his wrappers. At least one wrapper remains. Thus the final formula is

result = n [bars count bought for money] + (n-1)[wrappers count except one wrapper] // (m-1)

harshitvit + 1 comment try out this

for(int a0 = 0; a0 < t; a0++){ int n; int c; int m; int num=0; cin >> n >> c >> m; int wrapper=0; num=n/c; wrapper=num; while(wrapper>=m) { num=num+(wrapper/m); wrapper=(wrapper/m)+(wrapper%m); } cout<<num<<endl; }

agamyaji3 + 0 comments can you please explain this ?

meethinsu414 + 0 comments right ans is 898

aniketnath283 + 0 comments simpliest solution for any language solved for all test cases https://www.hackerrank.com/challenges/chocolate-feast/forum/comments/484679

chay_29 + 0 comments me too. only test case5

deepeshNair + 0 comments [deleted]deepeshNair + 0 comments You can check this out

https://www.hackerrank.com/challenges/chocolate-feast/forum/comments/561503

ameybhutada + 0 comments public class Solution {

`// Complete the chocolateFeast function below. static int chocolateFeast(int n, int c, int m) { int count=0; count=n/c; int chocolate=count; while(chocolate>=m) { chocolate=chocolate-m+1; count++; } return count; }`

icedtrees + 2 comments I came up with a nice formula for solving the problem.

We start off with a self-referential formula:

`t = money + wrappers t = floor(n/c) + floor((t-1)/m)`

The reason we use

`t-1`

over`t`

is because the wrapper you acquire from your last exchanged chocolate is not part of your available resources.We then solve for t:

`t - floor((t - 1) / m) = floor(n/c) ceil(t - (t - 1) / m) = floor(n/c) ceil((mt - t + 1) / m) = floor(n/c)`

We can convert this to an inequality. The idea is that

`mt - t + 1`

must be at between`floor(n/c)-1`

(non-inclusive) and`floor(n/c)`

(inclusive) multiples of`m`

.`m(floor(n/c) - 1) < t(m - 1) + 1 <= m(floor(n/c)) m(floor(n/c) - 1) - 1 < t(m - 1) <= m(floor(n/c)) - 1 (m(floor(n/c) - 1) - 1) / (m - 1) < t <= (m(floor(n/c)) - 1 ) / (m - 1)`

Note there can be multiple values of

`t`

that are within this range. We select the largest, because we want the maximum amount of chocolates`t = floor((m * floor(n/c) - 1) / (m - 1))`

Another form of this is what @ansonete described below:

`t = floor((m * floor(n/c) - floor(n/c) + floor(n/c) - 1) / (m - 1)) = floor(floor(n/c) + (floor(n/c) - 1) / (m - 1)) = floor(n/c) + (floor(n/c) - 1) / (m - 1)`

jjlearman + 0 comments The first line is wrong:

`t = money + wrappers`

It should be

`t = chocolates + wrappers`

ansonete + 9 comments I've found out that there is a direct solution (i didn't get it myself, I just found out by searching, I solved the problem with while bucle like almost everybody).

### Code

`money = 10 cost = 2 extra = 2 boughtCandies = money / cost result = boughtCandies + (boughtCandies - 1) / (extra - 1)`

Does anyone know the math basis or where this comes from, or where can I read about this?

Thanks!

abhiranjan + 0 comments @ ansonete, it's recommended to write formula instead of code snippet. Will be easier for non-python programmar :)

ansonete + 0 comments Ok, I updated it, with the input example 10 2 2 and i tried to remove the python stuff and leave only the math symbols :) If anyone has a suggestion how to better rewrite the question is welcome :)

zekedroid + 4 comments Well, not sure about this formula but if you wanted to use one yourself, think about the steps you would take if you were doing the exchange of wrappers by hand; by the way, not the best code to write as it's confusing and you are writing over the

`money`

variable instead of creating a new one. First you buy as many chocolates as you want,`chocolates = money/cost`

. Now you eat those chocolates and exchange all the wrappers possible for extra ones, let's call it`extras = chocolates / extra`

.BUT you could have leftover

`chocolates`

so let's see how many,`remainders = chocolates % extra`

. Since we got`extras`

, we can add it to the`remainders`

and see if it is enough for more chocolates,`super_extras = (extras + remainders) / extra`

. Remember that all of this is integer division so it'll round down.Finish by adding to the total

`chocolates += extras + super_extras`

and you're done! You have to tweak it a bit to get edge cases but that part should be trivial. Use the same logic to see where those terms you wrote came from. I think if you rearrange what I showed you'll get the same thing.ganuraag31 + 1 comment For just one test case input() There needs changing in condition. What if super_extras is more than m. Then more candies.

ssaurabh44 + 0 comments Chk this :Worked for all test cases: for(int a1=0;a1

`no = n/c; int wrapper = no; int rem = 0; while(wrapper>=m){ rem = wrapper/m; no = no +rem; rem = rem + wrapper%m; wrapper = rem; } System.out.println(no); }`

ansonete + 0 comments Thanks for your comment. You are right, the code is confusing due the name of the variables :) I updated it.

lawless_war1ock + 0 comments I am not able to figure out the mistake in this code any help wud be appreciated

# include

# include

# include

# include

int main() { long int a,b,c,t,d,e; scanf("%ld",&t); while(t--) { scanf("%ld %ld %ld",&a,&b,&c); e=a/b; d=e;

`long int count=0; while(e>=c) { e=e-c; count++; e=e+count; } printf("%ld",(count+d)); printf("\n"); }`

return 0; }

ka1gu0 + 1 comment Yeah! You are right and the description is easy. I sloved this problem with the same way. The following is my AC code :)

`int main() { /* Enter your code here. Read input from STDIN. Print output to STDOUT */ int t,n,c,m; cin>>t; while(t--){ cin>>n>>c>>m; int answer=0; // Computer answer(the total of chocolates Bob eats) // answer consists of two parts: directly buy with money and exchange with wrapper int answer1 = n / c; // the number of buying with money answer += answer1; int answer2 = 0; // the number of exchanging with wrapper int wrapper = answer1 / m; // the number of first exchange with wrapper int remain = answer1 % m; // the number of first remain wrapper answer2 += wrapper; int available = wrapper + remain; while (available / m != 0) { wrapper = available / m; remain = available % m; answer2 += wrapper; available = wrapper + remain; } answer += answer2; cout<<answer<<endl; } return 0; }`

gabrielptp + 0 comments Similar approach using python:

t = int(raw_input().strip()) for a0 in xrange(t): n,c,m = raw_input().strip().split(' ') n,c,m = [int(n),int(c),int(m)]

`result = 0 ch = n / c result +=ch while True: ch_w = ch / m result += ch_w ch_r = ch % m ch = ch_w + ch_r if ch < m: break print result`

fizzixio + 0 comments I too, would also like to know about this.

iiiiii + 0 comments thanks for the solution, it reminds me of the meme "code works: no idea why"

singhaniya + 0 comments sir if you the logic behind it then pls help me to understand that thank you

anna_earwen + 5 comments @ansonete: This is indeed the cleanest solution, and the logic is simple: for every m wrappers you get a candy, candy = wrapper, i.e. for every m wrappers you get 1 wrapper back. Thus, the actual "wrapper price" of one candy is (m - 1) rather than m. Now, dividing your original number of wrappers by the "reduced price" is unfair - you still have to pay "full price" (i.e. full m) for the very first "exchange candy". Thus, let us first put m aside to make sure we get our first free candy. So the amount of wrappers to divide by reduced price is (boughtCandy - m). Now, the total number of candies is:

`boughtCandy = money / cost`

`total = boughtCandy + 1 + (boughtCandy - m) / (m - 1)`

where 1 is the candy we bought for the m we put aside in the first place. This can be simplified as follows:

`total = boughtCandy + (m - 1)/(m - 1) + (boughtCandy - m) / (m - 1)`

`= boughtCandy + (m - 1 + boughtCandy - m) / ( m - 1 )`

`= boughtCandy + (boughtCandy - 1)/(m - 1)`

Et voila!

anderustarroz + 0 comments That was brilliant Anna, thanks ;)

bhushanz + 0 comments thats wrong bud... check for 2nd case of 1st test case

ansonete + 0 comments Awesome explanation. Thanks a lot.

Tejasduseja + 1 comment My idea was same as you but i am getting runtime error for test case 5.could u help me?

ROBOPRO + 1 comment int t,n,m,c; int wrappers,count;

`scanf("%d",t);`

for(int i =0;i

`wrappers = n/c; count = wrappers; while (wrappers >= m) { wrappers = wrappers - m; count = count + 1; wrappers = wrappers + 1; } printf("%d",count);`

}

/*The main logic is above u can compare it easily */

kyran + 0 comments [deleted]

akshay_sharma96 + 0 comments Amazing :)

rahultoraskarr + 1 comment Here you can find formula with the proof http://www.geeksforgeeks.org/g-fact-42/

kujicoso + 0 comments I'm trying to understand how this problem can be mapped to a n-ary tree.

I've tried several approaches but couldn't succeed.

Could you explain the approach you have in mind?

Thanks!

rohitpal210 + 0 comments Here's one way

int chocolateFeast(int n, int c, int m) { int bars,w; bars=n/c; if(bars<m) { return bars; } else if(bars>m) { w=bars; while(w>=m) { bars+=w/m; w=w%m+w/m; } } else bars+=1; return bars; }

akshar36 + 0 comments Very simple. 100% all test cases passed

static int chocolateFeast(int n, int c, int m) { int wrapper=0; wrapper=n/c; int choc_count=wrapper; while(wrapper>=m) { wrapper=wrapper-m;//exchange wrapper++;//new choc wrapper.. so ++ choc_count++;//increase choc count } return choc_count; }

xingyuyanebay + 2 comments The test case: 12 4 4 result is 3.

Why can't I borrow a Wrapper and when I finish mine, I return the Wrapper?abhiranjan + 0 comments No loans, please :)

jeswinaugustine + 1 comment that cracked me up.. cheers!! :P

anishaaanifa1112 + 0 comments cheers!!!!!

sandrobrayen + 1 comment My solution with javascript passed all test

function chocolateFeast(n, c, m) { let jumlahWrapper = Math.floor(n / c); let eaten = jumlahWrapper; while ((jumlahWrapper / m) >= 1) { let wrapp = Math.floor(jumlahWrapper / m); eaten += wrapp; jumlahWrapper = wrapp + (jumlahWrapper % m); } return eaten; }

akramrizwan8 + 0 comments nice Solution

giancluciano + 2 comments Short Python solution with lambda function

#!/bin/python3 import sys function = lambda n,c,m : n//c + 1/(m-1) * (n/c) - 1/(m-1) t = int(input().strip()) for a0 in range(t): n,c,m = input().strip().split(' ') n,c,m = [int(n),int(c),int(m)] print(int(function(n,c,m)))

nei_oliveira + 0 comments Amazing, lad. Congrats. You really showed them how it's done.

shravil_potdar + 1 comment why are you deviding by m-1? Please could you explain me this code in breif...

felds + 0 comments ### [Off topic]

Can we start using markdown for good? Enough of huge “titles”, unformatted code and this:

# include

# include

# include

# include

**If you know what I mean:**- Refuse to answer to questions with bad formatting: it's not your responsibility to understand other peoples mess.

Ask gently for them to fix the formatting (or at least setting the right code highlighting). - Start educating your peers;

**If you don't:**- Learn about the markdown basics;
- Learn about how to format your code in markdown;
- Alternatively, take a look at the pale
*Cheatsheet*link under the post box. It's not complete but it's better than nothing; - Start paying attention the the post preview under the textarea;
- Teach others and pass the word :)

</rant>

- Refuse to answer to questions with bad formatting: it's not your responsibility to understand other peoples mess.

chronokitsune321 + 1 comment I'm confused... The sample states that we get chocolates per trip, and that means we get wrappers per trip. However, this also means that we get an extra chocolates per trip, right?

Given 10 dollars with chocolates for 2 dollars each, you can get 5 chocolates. That's 5 wrappers. To get a free chocolate, you need 5 wrappers, which we have, so now we have 6 chocolates. This means we now have 1 extra wrapper (total: 6, extra: 1)

Next, we have 12 dollars with chocolates for 4 dollars each. That means we can purchase 3 chocolates, meaning we now have 3 wrappers, plus the extra 1 from the last purchase. We need 4 wrappers to get a free chocolate, which we have, so we get a free chocolate and consequently have 1 extra wrapper (total: 4, extra: 1)

Finally, we have 6 dollars to spend on chocolates for 2 dollars each. This means we can buy 3 chocolates, giving us 3 new wrappers, plus the extra 1 from the last purchase. We need 2 wrappers to obtain a free chocolate, and we actually have 4 wrappers, meaning we get 2 free chocolates and 2 more wrappers as a consequence. We can obtain 1 chocolate more for free as a result, leading to 1 final chocolate and 1 extra wrapper (total: 6, extra: 1)

So how is the sample output 6, 3, 5 when we can clearly obtain 6, 4, 6?

What am I missing?

enzo_borgfrantz + 1 comment I am having the same issue, not sure when the extra wrappers count and when they don't.

chronokitsune321 + 0 comments The way it works is the number of leftover wrappers doesn't carry over to the next trip to the store, meaning once you can no longer trade for a free chocolate, any extra wrappers you have are no longer able to be traded.

- You purchase 5 chocolates, meaning you have 5 wrappers. You trade the 5 wrappers for 1 free chocolate for a total of 6 chocolates. The wrapper for the 1 free chocolate is the only wrapper left. Because you have 1 wrapper, not 5, you cannot receive any more chocolate, so the remaining wrapper is discarded. Chocolates=5+1=6, extra=1
- You purchase 3 chocolates, meaning you have 3 wrappers. However, 4 wrappers are needed to trade for 1 free chocolate, so you cannot receive any chocolates. Those 3 wrappers are discarded. Chocolates=3, extra=3
- You purchase 3 chocolates, meaning you have 3 wrappers. 2 wrappers are needed to receive a free chocolate, so you trade in 2/3 of your wrappers for 1 free chocolate. This leaves you with 1 wrapper from the chocolate you paid for and 1 wrapper from the 1 free chocolate, or 2 wrappers total. You trade them both for 1 more chocolate, leaving you with 1 wrapper to be discarded. Chocolates=3+1+1=5, extra=1

sricodes + 0 comments Visualise this problem as full length node string . In each step M joins hand. and left outs are added in next strings .:) Good luck .. Took me a while but finally ended up in a small while loop.

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