Computer Science, asked by mmani35642, 10 months ago


Philaland Coin
- 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.​

Answers

Answered by mayank90345
77

Answer:#include <iostream>

using namespace std;

int main() {

   int t,den,a,ct;

   cin>>t;

   while(t--){

       cin>>den;

       a=1,ct=0;

       while(den>0){

           den-=a;

           a=a+1;

           ct++;

       }

       cout<<ct<<"\n";

   }

   

}

Explanation:first while will iterate for test cases then we will subtract our denomination till it passes 0.

Answered by poojan
1

Language using :- Python Programming

Program :-

t=int(input())

if t>=1 and t<=100:

  while(t>0):

      sum=0

      total=0

      n=int(input())

      if n>=1 and n<=5000:

          l=list(range(1,n+1))

          for i in l:

              if sum!=n+1 and sum<n+1:

                  sum=sum+i

                  total=total+1

                  if sum>n+1 or sum==n:

                      sum=sum-i

                      total=total-1

              if sum==n+1:

                  print(total)

                  break

      t=t-1

Input :-

2

10

5

Output :-

4

3

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