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: 140.00Input: 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++
// C++ program for above approach#include <iostream>using namespace std;// Function to return temperature// in the second thermometerdouble temp_convert(int F1, int B1, int F2, int B2, int T){ float t2; // Calculate the temperature t2 = F2 + (float)(B2 - F2) / (B1 - F1) * (T - F1); return t2;}// Driver Codeint 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;}// This code is contributed by kirti |
C
// C program for above approach#include <stdio.h>// Function to return temperature// in the second thermometerdouble temp_convert(int F1, int B1, int F2, int B2, int T){ float t2; // Calculate the temperature t2 = F2 + (float)(B2 - F2) / (B1 - F1) * (T - F1); return t2;}// Driver Codeint 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
// Java program for above approachimport java.io.*;class GFG{// Function to return temperature// in the second thermometerstatic double temp_convert(int F1, int B1, int F2, int B2, int T){ float t2; // Calculate the temperature t2 = F2 + (float)(B2 - F2) / (B1 - F1) * (T - F1); return t2;}// Driver Codepublic 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));}}// This code is contributed by rishavmahato348 |
Python3
# Python3 program for above approach# Function to return temperature# in the second thermometerdef temp_convert(F1, B1, F2, B2, T): # Calculate the temperature t2 = F2 + ((float)(B2 - F2) / (B1 - F1) * (T - F1)) return t2# Driver CodeF1 = 0B1 = 100F2 = 32B2 = 212T = 37print(temp_convert(F1, B1, F2, B2, T))# This code is contributed by Ankita Saini |
C#
// C# program for above approachusing System;class GFG{// Function to return temperature// in the second thermometerstatic double temp_convert(int F1, int B1, int F2, int B2, int T){ float t2; // Calculate the temperature t2 = F2 + (float)(B2 - F2) / (B1 - F1) * (T - F1); return t2;}// Driver Codepublic static void Main(){ int F1 = 0, B1 = 100; int F2 = 32, B2 = 212; int T = 37; //float t2; Console.Write(String.Format("{0:0.##}", temp_convert(F1, B1, F2, B2, T)));}}// This code is contributed by subhammahato348 |
Javascript
<script>// JavaScript program for above approach// Function to return temperature// in the second thermometerfunction temp_convert(F1, B1, F2, B2, T){ var t2; // Calculate the temperature t2 = F2 + (B2 - F2) / (B1 - F1) * (T - F1); return t2;}// Driver Codevar 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));// This code is contributed by Khushboogoyal499</script> |
98.60
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.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
