Convert a given temperature to another system based on given boiling and freezing points
Last Updated :
04 Jun, 2022
There are two thermometers and given five integers F1, B1, F2, B2, and T, where F1 and B1 are the Freezing points and Boiling Point of water on thermometer 1, and F2 and B2 are the Freezing points and Boiling Point of water on thermometer 2 respectively, and T is some temperature recorded on thermometer 1. The task is to find the temperature on the thermometer 2.
Example:
Input: F1 = 0, B1 = 10, F2 = 100, B2 = 200, T = 4
Output: 140.00
Input: F1 = 0, B1 = 100, F2 = 32, B2 = 212, T = 37
Output: 98.60
Approach: Consider the first thermometer to use U1 unit system and second thermometer using U2 unit system.
- The idea is to get the difference between the boiling and freezing point of water on each thermometer.
- The number of units between the freezing and boiling points of both thermometers shows the same temperature difference.
So, (B1 – F1) U1 == (B2 – F2) U2
By Unitary method, U1 = ( B2 – F2 ) / ( B1 – F1 ) U2
Relative value of U2 is T – F1 and U1 is T – F2
Hence, T = F2 + ( ( B2 – F2 ) / ( B1 – F1 ) ) * ( T – F1 )
Below is the implementation of the above approach:
C++
#include <iostream>
using namespace std;
double temp_convert( int F1, int B1, int F2,
int B2, int T)
{
float t2;
t2 = F2 + ( float )(B2 - F2) /
(B1 - F1) *
(T - F1);
return t2;
}
int main()
{
int F1 = 0, B1 = 100;
int F2 = 32, B2 = 212;
int T = 37;
float t2;
cout << temp_convert(F1, B1, F2, B2, T);
return 0;
}
|
C
#include <stdio.h>
double temp_convert( int F1, int B1, int F2,
int B2, int T)
{
float t2;
t2 = F2
+ ( float )(B2 - F2)
/ (B1 - F1) * (T - F1);
return t2;
}
int main()
{
int F1 = 0, B1 = 100;
int F2 = 32, B2 = 212;
int T = 37;
float t2;
printf ( "%.2f" ,
temp_convert(F1, B1, F2, B2, T));
return 0;
}
|
Java
import java.io.*;
class GFG{
static double temp_convert( int F1, int B1, int F2,
int B2, int T)
{
float t2;
t2 = F2 + ( float )(B2 - F2) /
(B1 - F1) *
(T - F1);
return t2;
}
public static void main(String[] args)
{
int F1 = 0 , B1 = 100 ;
int F2 = 32 , B2 = 212 ;
int T = 37 ;
float t2;
System.out.printf( "%.2f" ,
temp_convert(F1, B1, F2, B2, T));
}
}
|
Python3
def temp_convert(F1, B1, F2, B2, T):
t2 = F2 + (( float )(B2 - F2) /
(B1 - F1) *
(T - F1))
return t2
F1 = 0
B1 = 100
F2 = 32
B2 = 212
T = 37
print (temp_convert(F1, B1, F2, B2, T))
|
C#
using System;
class GFG{
static double temp_convert( int F1, int B1, int F2,
int B2, int T)
{
float t2;
t2 = F2 + ( float )(B2 - F2) /
(B1 - F1) *
(T - F1);
return t2;
}
public static void Main()
{
int F1 = 0, B1 = 100;
int F2 = 32, B2 = 212;
int T = 37;
Console.Write(String.Format( "{0:0.##}" ,
temp_convert(F1, B1, F2, B2, T)));
}
}
|
Javascript
<script>
function temp_convert(F1, B1, F2, B2, T)
{
var t2;
t2 = F2 + (B2 - F2) /
(B1 - F1) *
(T - F1);
return t2;
}
var F1 = 0, B1 = 100;
var F2 = 32, B2 = 212;
var T = 37;
var t2;
document.write(temp_convert(
F1, B1, F2, B2, T).toFixed(2));
</script>
|
Time Complexity: O(1), since there is no loop and basic arithmetic is performed.
Auxiliary Space: O(1), since no extra space has been taken.
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