 # spoj MAIN8_C - Shake Shake Shaky.

Shaky has N (1<=N<=50000) candy boxes each of them contains a non-zero number of candies (between 1 and 1000000000). Shakey want to distibute these candies among his K (1<=K<=1000000000) IIIT-Delhi students. He want to distibute them in a way such that:

1. All students get equal number of candies.

2. All the candies which a student get must be from a single box only.

As he want to make all of them happy so he want to give as many candies as possible. Help Shaky in finding out what is the maximum number of candies which a student can get.

Input
First line contains 1<=T<=20 the number of test cases. Then T test cases follow. First line of each test case contains N and K. Next line contains N integers, ith of which is the number of candies in ith box.

Output
For each test case print the required answer in a seperate line.

Example
Input:
2
3 2
3 1 4
4 1
3 2 3 9

Output:
3
9

Someone please explain how to use binary search here.

It’s a good example of modified binary search.

Here is the approach:

1.Sort the array in ascending order.

2.Run the binary search and Let say mid element is x.

3.Now make a check whether that mid element fulfills your demand according to a number of children.

4.If condition satisfies, then move b/w mid and high else move b/w low and mid.

Here is the code after some modification in binary search:https://pastebin.com/4n86mGQn

EDIT: Explanation of sample test case:

``````Given array: 3 1 4
After sorting: 1 3 4
N = 3 & K = 2;
1st iteration: mid = (0+3)/2 = 1;
a[mid] = 3 = val;
Now make a check whether val satisfies our condition or not?
Here is the function for the same:

bool func(ll val,ll k,ll n)
{
if(k==1)
return true;
ll sum=0;
for(ll i=n-1;i>=0;i--)
{
sum +=(a[i]/val);
}
if(sum >= k)
return true;
else
return false;
}
``````

Here is the code explanation:
As we looping from N-1 to 0,our array becomes `4 3 1`, so not to check whether one/more child can get candy out of box or not,`sum = sum+a[i]/val;`so now

``````4 -> 4/3 = 1
3 -> 3/3 = 1
1 -> 1/3 = 0
sum = 2 = k,return true;
``````

Happy coding!

3.Now make a check whether that mid element fulfills your demand according to a number of children.

3 2
3 1 4
lets say n=4, k(no of child)=2.

man the test case. n=3,k=2. values of n=3,1,4. explain this

for(ll i=n-1;i>=0;i–)
{
sum +=(a[i]/val);
}

explain this too

oh, my bad didn’t notice that.

ayush please explain me the condition. i cannot fully understand this question

try run your sample test case, you will understand that.Sorry can’t give more time to this problem, I have other stuff to do.

sum +=(a[i]/val);. does this mean number of students who got chocolates?

Yupppppppppp

``````    lo=0;
hi=1000000000;
while(lo<=hi)
{

mid=(lo+hi)/2;
if(mid==0)
{
if(k>n) ans=0;
else ans=1;
break;
}
stud=0;
for(i=0;i<n;i++)
{
stud+=a[i]/mid;
}
if(stud>=k)
{
ans=mid;
lo=mid+1;
}
else hi=mid-1;
}
cout<<ans<<endl;
}
return 0;
``````

}

can anyone explain why I’m getting WA!
https://pastebin.com/gaymS01m

Change line 24 high=mid
bcoz for each iteration we are checking [0,high) as first:high=arr[n-1]+1
be careful high is excluded

//