Question

Problem Description The problem solvers have found a new Island for coding and named it as Philaland. These smart people were given a task to make purchase of items at the Island easier by distributing various coins with different value Manish has come up with a solution that if we make coins category starting from $1 till the maximum price of item present on Island, then following example to prove his point. Lets suppose the maximum price of an item is 5$ then we can make coins of ($1, 52, 53, 54, 55) to purchase any item ranging from $1 till 55. Now Manisha, being a keen observer suggested that we could actually minimize the number of coins required and gave following distribution ( be purchased one time ranging from $1 to $5. Everyone was impressed with both of them. Your task is to help Manisha come up with minimum number of denominations for any arbitrary max price in Philaland. Constraints 1<=T<=100 1<=N<=5000 Input Format First line contains an integer T denoting the number of test cases. Next T lines contains an integer N denoting the maximum price of the item present on Philaland.

Answer 1

Philaland Coin Problem

Language used : Python Programming

Program :

no_of_testcases=int(input())

l=[]

for i in range(no_of_testcases):

   l.append(int(input()))

for i in l:

   count=0

   while i>=1:

       i=i//2

       count=count+1

   print(count)

Input :

2

10

5

Output :

4

3

Explanation :

• If the amount needed = 10, minimum no.of coins needed = {1,2,3,4}

       If the amount needed = 5, minimum no.of coins needed = {1,2,3}

• If we see, the coin count is close to the nearest 2's power + 1.

• The denomination goes as 10 on entering the loop checks if it is greater than or equal to 1. Yes? Now, i becomes 5 and count increases by 1.

• Next 5>=1 is True, i=5//2 which is 2 that now stores in i, making count 2

• 2>=1, true. i=2//2 and the value 1 is stored in i and count becomes 3

• 1>=1, true. i=1//2 and the result 0 stores in i, count becomes 4.

• 0>=1, False. Loop terminates and count=4 prints. Similarly, the computation happens with 5.

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