## 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-

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