Fill Candies---Difficulty: 681---Python Solution

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FILLCANDIES-EDITORIAL

PROBLEM LINK:

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Setter: soumyadeep_21
Testers: tabr, tejas10p
Editorialist: kiran8268

DIFFICULTY:

681

PREREQUISITES:

None

PROBLEM:

The objective is to find the minimum number of bags Chef needs so that he can put every candy into bags, where each bag has K pockets and each pocket can have a maximum of M candies.

EXPLANATION:

Given each bag has K pockets which can have maximum of M candies, the maximum number of candies that a bag can hold is K\times M.
Thus if:

  • N<(K\times M), the minimum number of bags required is 1.
    *N%(K\times M)=0,the minimum number of bags required is N \div (K\times M)
    *N%(K\times M)>0,the minimum number of bags required is [N \div (K\times M) ]+1

TIME COMPLEXITY:

Time complexity is O(1).

SOLUTION:

My Solution

t = int(input())
for _ in range(t):
    n, k ,m = map(int, input().split())
    if n < (k*m):
        print(1)
    elif n%(k*m) == 0:
        print(n//(k*m))
    else:
        print((n//(k*m))+1)

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3 4 / 4

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