Chef and Division 3 Codechef January Challenge 2021 Division 3

Chef and Division 3 | Problem Code: DIVTHREE

Solved

chef-and-division-3-codechef-january-challenge-2021-division-3

Question Link : https://www.codechef.com/JAN21C/problems/DIVTHREE

Video Link : https://www.youtube.com/watch?v=ItN6GMcuqmA

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 1≤i≤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 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

  • 1≤T≤103
  • 1≤N≤102
  • 1≤K≤109
  • 1≤D≤109
  • 1≤Ai≤107 for each valid i

Subtasks

Subtask #1 (40 points):

  • N=1
  • 1≤A1≤105

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.

Solution Code : IN C++

#include <bits/stdc++.h>  // This will work only for g++ compiler. 
#define for0(i, n) for (int i = 0; i < (int)(n); ++i) // 0 based indexing
#define for1(i, n) for (int i = 1; i <= (int)(n); ++i) // 1 based indexing
#define forc(i, l, r) for (int i = (int)(l); i <= (int)(r); ++i) // closed interver from l to r r inclusive
#define forr0(i, n) for (int i = (int)(n) - 1; i >= 0; --i) // reverse 0 based.
#define forr1(i, n) for (int i = (int)(n); i >= 1; --i) // reverse 1 based

//short hand for usual tokens
#define pb push_back
#define fi first
#define se second

// to be used with algorithms that processes a container Eg: find(all(c),42)
#define all(x) (x).begin(), (x).end() //Forward traversal
#define rall(x) (x).rbegin, (x).rend() //reverse traversal

// traversal function to avoid long template definition. Now with C++11 auto alleviates the pain.
#define tr(c,i) for(__typeof__((c)).begin() i = (c).begin(); i != (c).end(); i++)

// find if a given value is present in a container. Container version. Runs in log(n) for set and map
#define present(c,x) ((c).find(x) != (c).end())

//find version works for all containers. This is present in std namespace.
#define cpresent(c,x) (find(all(c),x) != (c).end())

// Avoiding wrap around of size()-1 where size is a unsigned int.
#define sz(a) int((a).size())

#define LL long long
using namespace std;

// Shorthand for commonly used types
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef pair<int, int> ii;
typedef vector<ii> vii;
typedef long long ll;
typedef vector<ll> vll;
typedef vector<vll> vvll;
typedef double ld;


int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    LL int n,t,k,d,a[100000],sum;
    cin>>t;
    while(t){
        cin>>n>>k>>d;
        for0(i, n) cin>>a[i];
        for0(i, n) sum+=a[i];
        int j =sum/k;
        if(j>=d) cout<<d<<"\n";
        else cout<<j<<"\n";
        sum=0;
        t--;
    }
    return 0;
}