Program to find the head start in a race
Last Updated :
31 Aug, 2022
Given the head start that A gives to B and C in a 100-meters race. The task is to find the head-start that B can give to C in the same race.
Examples:
Input: B = 10 meters, C = 28 meters
Output: 20 meters
B can give C a start of 20 meters.
Input: B = 20 meters, C = 50 meters
Output: 62 meters
B can give C a start of 62 meters.
Approach:
Total meters in a race = 100 meters.
A is ahead of B by 10 meters. When A completed it’s 100 meters B completed it’s 90 meters.
Similarly, A is ahead of C by 28 meters. When A completed it’s 100 meters C completed it’s 72 meters.
Now, When B completed it’s 90 meters C completed it’s 72 meters.
So when B completed it’s 100 meters C completed it’s 80 meters.
–> (( C * 100) / B)
–> (( 72 * 100) / 90) i.e 80 meters
So B can give C a start of 20 meters
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
int Race( int B, int C)
{
int result = 0;
result = ((C * 100) / B);
return 100 - result;
}
int main()
{
int B = 10, C = 28;
B = 100 - B;
C = 100 - C;
cout << Race(B, C) << " meters" ;
return 0;
}
|
Java
public class GFG
{
static int Race( int B, int C)
{
int result = 0 ;
result = ((C * 100 ) / B);
return 100 - result;
}
public static void main(String[] args)
{
int B = 10 ;
int C = 28 ;
B = 100 - B;
C = 100 - C;
System.out.println(Race(B, C) + " meters" );
}
}
|
Python3
def Race(B, C):
result = 0 ;
result = ((C * 100 ) / / B)
return 100 - result
if __name__ = = "__main__" :
B = 10
C = 28
B = 100 - B;
C = 100 - C;
print ( str (Race(B, C)) + " meters" )
|
C#
using System;
class GFG
{
static int Race( int B, int C)
{
int result = 0;
result = ((C * 100) / B);
return 100 - result;
}
public static void Main()
{
int B = 10;
int C = 28;
B = 100 - B;
C = 100 - C;
Console.Write(Race(B, C) + " meters" );
}
}
|
PHP
<?php
function Race( $B , $C )
{
$result = 0;
$result = (( $C * 100) / $B );
return 100 - $result ;
}
$B = 10;
$C = 28;
$B = 100 - $B ;
$C = 100 - $C ;
echo Race( $B , $C ) . " meters" ;
?>
|
Javascript
<script>
function Race(B, C)
{
var result = 0;
result = ((C * 100) / B);
return 100 - result;
}
var B = 10, C = 28;
B = 100 - B;
C = 100 - C;
document.write(Race(B, C) + " meters" );
</script>
|
Time Complexity: O(1)
Auxiliary Space: O(1)
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