Marathon CodeChef Solution

Problem – Marathon CodeChef Solution

Chefland is holding a virtual marathon for the categories 10 km, 21 km and 42 km having prizes A,B,C (A<B<C) respectively to promote physical fitness. In order to claim the prize in a particular category the person must cover the total distance for that category within D days. Also a single person cannot apply in multiple categories.

Given the maximum distance d km that Chef can cover in a single day, find the maximum prize that Chef can win at the end of D days. If Chef can’t win any prize, print 0.

###Input

• The first line contains an integer T, the number of test cases. Then the test cases follow.
• Each test case contains a single line of input, five integers D,d,A,B,C.

###Output For each test case, output in a single line the answer to the problem.

###Constraints

• 1≤T≤50
• 1≤D≤10
• 1≤d≤5
• 1≤A<B<C≤10^5

Sample 1:

Input:
3
1 1 1 2 3
10 1 1 2 3
10 3 1 2 3
Output:
0
1
2

Explanation:

Test Case 1: The maximum distance covered by Chef is 1⋅1=1 km which is less than any of the available distance categories. Hence Chef won’t be able to claim a prize in any of the categories.

Test Case 2: The maximum distance covered by Chef is 10⋅1=10 km which is equal to distance of the first category. Hence Chef can claim the maximum prize of 1 units.

Test Case 3: The maximum distance covered by Chef is 10⋅3=30 km which is more than the distance of the second category but less than that of the third category. Hence Chef can claim the maximum prize of 2 units.

Marathon CodeChef Solution in Java

/* package codechef; // don't place package name! */

import java.util.*;
import java.lang.*;
import java.io.*;

/* Name of the class has to be "Main" only if the class is public. */
class Codechef
{
	public static void main (String[] args) throws java.lang.Exception
	{
		Scanner in = new Scanner(System.in);
		int t = in.nextInt();
		
		while(t-- >0){
		    int D = in.nextInt();
		    int d = in.nextInt();
		    int a = in.nextInt();
		    int b = in.nextInt();
		    int c = in.nextInt();
		    
		    int ans = (d*D);
		    if(ans>=0 &&ans<10){
		        System.out.println(0);
		    }else if(ans<21){
		        System.out.println(a);
		    }else if (ans<42) {
		        System.out.println(b);
		    }else{
		        System.out.println(c);
		    }
		}
	}
}

Marathon CodeChef Solution in Pyth 3

test_cases = int(input(" "))

for i in range (0,test_cases):
    text = input(" ")
    (D,d,a,b,c) = map(int,text.split(" "))
    ran = D * d
    if ran >= 0 and ran < 10:
        print(0)
    elif ran < 21:
        print(a)
    elif ran < 42:
        print(b)
    else:
        print(c)


                
    

Marathon CodeChef Solution in C++17

#include <iostream>
using namespace std;

int main() {
	// your code goes here
	int t;
	cin>>t;
	while(t--)
	{
	    int D,d,A,B,C;
	    cin>>D>>d>>A>>B>>C;
	    
	    if(d*D>=42)
	    cout<<C<<endl;
	    else if(d*D>=21)
	    cout<<B<<endl;
	    else if(d*D>=10)
	    cout<<A<<endl;
	    else
	    cout<<0<<endl;
	}
	return 0;
}

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