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Chef and Division 3 DIVTHREE SOLUTION Code Chef
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.Chef and Division 3 DIVTHREE SOLUTION
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.Chef and Division 3 DIVTHREE SOLUTION
Solution
Program Python:
try: T=int(input()) for _ in range(T): N,K,D = map(int,input().split()) arr = list(map(int,input().split())) if(sum(arr)<K): print(0) else: res=sum(arr)//K if res>D: print(D) else: print(res) except: pass
Program C++:
#include <iostream> using namespace std; int main() { // your code goes here int t; cin>>t; while(t--) { int n; long k,d; cin>>n>>k>>d; long long total = 0; long A[n] = { 0 }; for(int i=0;i<n;i++) { cin>>A[i]; total += A[i]; } // long long max = k * d; long days = total / k; if (days > d) { cout<<d<<endl; } else { cout<<days<<endl; } } return 0; }
Program Java:
import java.util.*; import java.lang.*; import java.io.*; class div3 { public static void main (String[] args) throws java.lang.Exception { try { Scanner scn=new Scanner(System.in); int t=scn.nextInt(); for(int l=0;l<t;l++) { int n=scn.nextInt(); int k=scn.nextInt(); int d=scn.nextInt(); int tprob=0; int a[]=new int[n]; for(int i=0;i<n;i++) { a[i]=scn.nextInt(); tprob+=a[i]; } int days=tprob/k; if(d<days) System.out.println(d); else { System.out.println(days); } } } catch(Exception e) {} } }
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