Skip to main content

Chef and Division 3 Codechef january challenge solution || Chef and Division 3 Codechef january challenge Editorial 2021

 Chef and Division 3  Codechef january challenge solution || Chef and Division 3  Codechef january challenge Editorial-

Problem statement-

Chef wants to host some Division-3 contests. Chef has N setters who are busy creating new problems for him. The ith setter has made Ai problems where 1iN.

A Division-3 contest should have exactly K problems. Chef wants to plan for the next D days using the problems that they have currently. But Chef cannot host more than one Division-3 contest in a day.

Given these constraints, can you help Chef find the maximum number of Division-3 contests that can be hosted in these D days?

Input:

  • The first line of input contains a single integer T denoting the number of test cases. The description of T test cases follows.
  • The first line of each test case contains three space-separated integers - NK and D respectively.
  • The second line of each test case contains N space-separated integers A1,A2,,AN respectively.

Output:

For each test case, print a single line containing one integer ― the maximum number of Division-3 contests Chef can host in these D days.

Constraints

  • 1T103
  • 1N102
  • 1K109
  • 1D109
  • 1Ai107 for each valid i

Subtasks

Subtask #1 (40 points):

  • N=1
  • 1A1105

Subtask #2 (60 points): Original constraints

Sample Input:

5
1 5 31
4
1 10 3
23
2 5 7
20 36
2 5 10
19 2
3 3 300
1 1 1

Sample Output:

0
2
7
4
1

Explanation:

  • Example case 1: Chef only has A1=4 problems and he needs K=5 problems for a Division-3 contest. So Chef won't be able to host any Division-3 contest in these 31 days. Hence the first output is 0.

  • Example case 2: Chef has A1=23 problems and he needs K=10 problems for a Division-3 contest. Chef can choose any 10+10=20 problems and host 2 Division-3 contests in these 3 days. Hence the second output is 2.

  • Example case 3: Chef has A1=20 problems from setter-1 and A2=36 problems from setter-2, and so has a total of 56 problems. Chef needs K=5 problems for each Division-3 contest. Hence Chef can prepare 11 Division-3 contests. But since we are planning only for the next D=7 days and Chef cannot host more than 1 contest in a day, Chef cannot host more than 7 contests. Hence the third output is 7.

Chef and Division 3  Codechef january challenge solution || Chef and Division 3  Codechef january challenge Editorial


Solution-

This problem can be solve using brute force approach.
calculate sum of array and print cout<<min(sum/k,d)<<endl;
here is my code-

Join telegram channel for code and editorial.-https://t.me/competitiveProgrammingDiscussion

Comments

Popular posts from this blog

codeforces rating system | Codeforces rating Newbie to Legendary Grandmaster

 Codeforces rating system | Codeforces rating Newbie to Legendary Grandmaster- Codeforces is one of the most popular platforms for competitive programmers and  codeforces rating matters a lot  .  Competitive Programming  teaches you to find the easiest solution in the quickest possible way. CP enhances your problem-solving and debugging skills giving you real-time fun. It's brain-sport. As you start solving harder and harder problems in live-contests your analytical and rational thinking intensifies. To have a good codeforces profile makes a good impression on the interviewer. If you have a good  codeforces profile so it is very easy to get a referral for product base company like amazon, google , facebook etc.So in this blog I have explained everything about codeforces rating system. What are different titles on codeforces- based on rating codeforces divide rating into 10 part. Newbie Pupil Specialist Expert Candidate Codemaster Master International Master Grandmaster Internat

Apple Division CSES Problem Set Solution | CSES Problem Set Solution Apple division with code

 Apple Division CSES Problem Set Solution | CSES Problem Set Solution Apple division with code - Apple Division CSES Problem Solution Easy Explanation. Apple division is problem is taken form cses introductory problem set.Let's Read Problem statement first. Problem Statement- Time limit:  1.00 s   Memory limit:  512 MB There are  n n  apples with known weights. Your task is to divide the apples into two groups so that the difference between the weights of the groups is minimal. Input The first input line has an integer  n n : the number of apples. The next line has  n n  integers  p 1 , p 2 , … , p n p 1 , p 2 , … , p n : the weight of each apple. Output Print one integer: the minimum difference between the weights of the groups. Constraints 1 ≤ n ≤ 20 1 ≤ n ≤ 20 1 ≤ p i ≤ 10 9 1 ≤ p i ≤ 10 9 Example Input: 5 3 2 7 4 1 Output: 1 Explanation: Group 1 has weights 2, 3 and 4 (total weight 9), and group 2 has weights 1 and 7 (total weight 8). Join Telegram channel for code discussi

Merge Two sorted array Without Extra Space

Problem statement-  Given two sorted arrays arr1[] and arr2[] of sizes N and M in non-decreasing order. Merge them in sorted order without using any extra space. Modify arr1 so that it contains the first N elements and modify arr2 so that it contains the last M elements.    Example 1: Input:  N = 4, arr1[] = [1 3 5 7]  M = 5, arr2[] = [0 2 6 8 9] Output:  arr1[] = [0 1 2 3] arr2[] = [5 6 7 8 9] Explanation: After merging the two  non-decreasing arrays, we get,  0 1 2 3 5 6 7 8 9.   Example 2: Input:  N = 2, arr1[] = [10, 12]  M = 3, arr2[] = [5 18 20] Output:  arr1[] = [5 10] arr2[] = [12 18 20] Explanation: After merging two sorted arrays  we get 5 10 12 18 20. Your Task: You don't need to read input or print anything. You only need to complete the function merge() that takes arr1, arr2, N and M as input parameters and modifies them in-place so that they look like the sorted merged array when concatenated. Expected Time Complexity:  O((n+m) log(n+m)) Expected Auxilliary Space: O(1