World Record Codechef Solution

World Record Codechef Solution

Problem

This news headline has been echoing in Chef’s mind since 2009 when he started his career in sprinting. This time in the 2021 Tokyo Olympics he is determined to make a new world record.

There are mainly 3 important factors that result in the variation in speed during practice and during competition.

  • The first factor is a difference in track material which results in variation in the speed at competition with respect to practice by a factor of k1.
  • The second factor is wind speed which results in variation in the speed at competition with respect to practice by a factor of k2.
  • The final factor is the maximum speed achieved during practice. If the maximum speed during practice is v m/s then in competition it will be k3∗v m/s.

Given Chef’s max speed v during practice and the factors k1,k2,k3, find whether he will be able to create a new world record, i.e, can he complete 100 m in less than 9.58 seconds?

Note: The final time displayed by the Olympics clock is showing after rounding the original time to 2 places of decimal and is the only measure of runners performance.

Input

  • First line will contain T, number of testcases. Then the testcases follow.
  • Each testcase contains of a single line of input, four floats k1,k2,k3,v.

Output

Output in a single line, the answer, which should be “YES” if it’s possible for Chef to create a new world record and “NO” if not.

You may print each character of the string in uppercase or lowercase (for example, the strings “yEs”, “yes”, “Yes” and “YES” will all be treated as identical).

Constraints

  • 1 <= T <= 10^5 
  • 0 < k1, k2 < 2
  • 0 < k3 <= 1
  • 9 < v < 11
  • k1, k2, k3 contain 1 digit after decimal.
  • v contains 2 digits after decimal.

Subtasks

Subtasks #1 (100 Points): Original Constraints

Sample Input:

3
1.0 1.0 1.0 10.45
1.0 1.0 1.0 10.44
1.0 1.0 0.9 10.44

Sample Output:

YES
NO
NO

Explanation:

TestCase 1: Final speed of Chef after considering all the factors will be 1∗1∗1∗10.45=10.45 m / sec. So the time taken to complete the race will be 100/10.45=9.569=9.57 sec after rounding to 2 places after decimal. Since the time is strictly less than the world record time, therefore Chef can break the record.

TestCase 2: Final speed of Chef after considering all the factors will be 1∗1∗1∗10.44=10.44 m / sec. So the time taken to complete the race will be 100/10.44=9.578=9.58 sec after rounding to 2 places after decimal. Since the time is equal to the world record time, therefore Chef can’t break the record.

TestCase 3: Final speed of Chef after considering all the factors will be 1∗1∗0.9∗10.44=9.396 m / sec. So the time taken to complete the race will be 100/9.396=10.643=10.64 sec after rounding to 2 places after decimal. Since the time is strictly more than the world record time, therefore Chef can’t break the record.

World Record – CodeChef Solution in Python

T = int(input())
for i in range(T):
    a, b, c, d = [float(speed) for speed in input().split()]
    speed = a * b * c * d
    speed = round(100 / speed, 2)
    if speed < 9.58:
        print("YES")
    else:
        print("NO")

World Record – CodeChef Solution in CPP

n>>k1>>k2>>k3>>v;
	    float var=100.00/(k1*k2*k3*v);
	    float k=9.575;
	    if (var>=k)
	    cout<<"No"<<endl;
	    else
	    cout<<"Yes"<<endl;
	}
	return 0;
}

World Record – CodeChef Solution in JAVA

import java.util.Scanner;
public class Main {
  public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);
    int T = sc.nextInt();
    for (int tc = 0; tc < T; ++tc) {
      double k1 = sc.nextDouble();
      double k2 = sc.nextDouble();
      double k3 = sc.nextDouble();
      double v = sc.nextDouble();
      System.out.println(solve(k1, k2, k3, v) ? "YES" : "NO");
    }
    sc.close();
  }
  static boolean solve(double k1, double k2, double k3, double v) {
    return Math.round(100 / k1 / k2 / k3 / v * 100) < 958;
  }
}

Disclaimer: The above Problem (World Record) is generated by CodeChef but the solution is provided by Chase2learn.This tutorial is only for Educational and Learning purpose.

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