# Program to find the head start in a race

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.
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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++

 `// C++ implementation of above approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to find the B start to C ` `int` `Race(``int` `B, ``int` `C) ` `{ ` `    ``int` `result = 0; ` ` `  `    ``// When B completed it's 100 meter ` `    ``// then Completed meters by C is ` `    ``result = ((C * 100) / B); ` ` `  `    ``return` `100 - result; ` `} ` ` `  `// Driver Code. ` `int` `main() ` `{ ` `    ``int` `B = 10, C = 28; ` ` `  `    ``// When A completed it's 100 meter ` `    ``// Then completed meters of B and C is ` `    ``B = 100 - B; ` `    ``C = 100 - C; ` ` `  `    ``cout << Race(B, C) << ``" meters"``; ` ` `  `return` `0; ` `} `

## Java

 `// Java implementation of above approach ` `public` `class` `GFG  ` `{ ` `     `  `// Function to find the B start to C ` `static` `int` `Race(``int` `B, ``int` `C) ` `{ ` `    ``int` `result = ``0``; ` ` `  `    ``// When B completed it's 100 meter ` `    ``// then Completed meters by C is ` `    ``result = ((C * ``100``) / B); ` ` `  `    ``return` `100` `- result; ` `} ` ` `  `// Driver Code ` `public` `static` `void` `main(String[] args)  ` `{ ` `    ``int` `B = ``10``; ` `    ``int` `C = ``28``; ` `     `  `    ``// When A completed it's 100 meter ` `    ``// Then completed meters of B and C is ` `    ``B = ``100` `- B; ` `    ``C = ``100` `- C; ` `     `  `    ``System.out.println(Race(B, C) + ``" meters"``); ` `} ` `} ` ` `  `// This code is contributed  ` `// by ChitraNayal `

## Python 3

 `# Python 3 implementation  ` `# of above approach ` ` `  `# Function to find the  ` `# B start to C ` `def` `Race(B, C): ` ` `  `    ``result ``=` `0``; ` ` `  `    ``# When B completed it's 100 meter ` `    ``# then Completed meters by C is ` `    ``result ``=` `((C ``*` `100``) ``/``/` `B) ` ` `  `    ``return` `100` `-` `result ` ` `  `# Driver Code ` `if` `__name__ ``=``=` `"__main__"``: ` `    ``B ``=` `10` `    ``C ``=` `28` `     `  `    ``# When A completed it's 100 meter ` `    ``# Then completed meters of B and C is ` `    ``B ``=` `100` `-` `B; ` `    ``C ``=` `100` `-` `C; ` `     `  `    ``print``(``str``(Race(B, C)) ``+` `" meters"``) ` ` `  `# This code is contributed  ` `# by ChitraNayal `

## C#

 `// C# implementation of above approach ` `using` `System; ` `class` `GFG  ` `{ ` ` `  `// Function to find the B start to C ` `static` `int` `Race(``int` `B, ``int` `C) ` `{ ` `    ``int` `result = 0; ` ` `  `    ``// When B completed it's 100 meter ` `    ``// then Completed meters by C is ` `    ``result = ((C * 100) / B); ` ` `  `    ``return` `100 - result; ` `} ` ` `  `// Driver Code ` `public` `static` `void` `Main()  ` `{ ` `    ``int` `B = 10; ` `    ``int` `C = 28; ` `     `  `    ``// When A completed it's 100 meter ` `    ``// Then completed meters of B and C is ` `    ``B = 100 - B; ` `    ``C = 100 - C; ` `     `  `    ``Console.Write(Race(B, C) + ``" meters"``); ` `} ` `} ` ` `  `// This code is contributed  ` `// by ChitraNayal `

## PHP

 ` `

Output:

```20 meters
```

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Improved By : chitranayal