## 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 ${i}^{th}$ setter has made ${A}_{i}$ problems where $1\le i\le N$.

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 - $N$$K$ and $D$ respectively.
• The second line of each test case contains $N$ space-separated integers ${A}_{1},{A}_{2},\dots ,{A}_{N}$ 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

• $1\le T\le {10}^{3}$
• $1\le N\le {10}^{2}$
• $1\le K\le {10}^{9}$
• $1\le D\le {10}^{9}$
• $1\le {A}_{i}\le {10}^{7}$ for each valid $i$

• $N=1$
• $1\le {A}_{1}\le {10}^{5}$

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 ${A}_{1}=4$ 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 ${A}_{1}=23$ 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 ${A}_{1}=20$ problems from setter-1 and ${A}_{2}=36$ 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 $7$.

### 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

### 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

### Movie Festival CSES problem set solution

Movie Festival CSES problem set solution - Problem statement-  Time limit:  1.00 s   Memory limit:  512 MB In a movie festival  n n  movies will be shown. You know the starting and ending time of each movie. What is the maximum number of movies you can watch entirely? Input The first input line has an integer  n n : the number of movies. After this, there are  n n  lines that describe the movies. Each line has two integers  a a  and  b b : the starting and ending times of a movie. Output Print one integer: the maximum number of movies. Constraints 1 ≤ n ≤ 2 ⋅ 10 5 1 ≤ n ≤ 2 ⋅ 10 5 1 ≤ a < b ≤ 10 9 1 ≤ a < b ≤ 10 9 Example Input: 3 3 5 4 9 5 8 Output: 2 Solution - Step -1 take input in a vector as a pair first element is ending time ans second element is starting time Step -2 sort the vector in increasing order because we will watch that movie which is ending first. Step -3 iterate over vector and calculate ans .           follow code below-