### Fahrenheit to Celsius

Level EASY

#### Given three values - Start Fahrenheit Value (S), End Fahrenheit value (E) and Step Size (W), you need to convert all Fahrenheit values from Start to End at the gap of W, into their corresponding Celsius values and print the table.

##### Input Format :

##### Output Format :

`Fahrenheit to Celsius conversion table. One line for every`

` `

Fahrenheit and corresponding Celsius value. On Fahrenheit value and its

`corresponding Celsius value should be separate by tab ("\t")`

##### Constraints :

##### Sample Input 1:

##### Sample Output 1:

##### Sample Input 2:

##### Sample Output 2:

##### Explanation For Input 2:

`We need need to start calculating the Celsius values for each of`

`the Fahrenheit Value which starts from 20.`

So starting from 20 which is the given Fahrenheit start value, we need

to compute its corresponding Celsius value which computes

to -6. We print this information as <Fahrenheit Value> a tab

space"\t" <Celsius Value> on each line for each step of 13

we take to get the next value of Fahrenheit and extend this

idea till we reach the end that is till 119 in this case.

You may or may not exactly land on the end value depending on the

steps you are taking.

Not happy with this solution

ReplyDeletethe constraints of s and w are not covered in this code

ReplyDeleteimport java.util.Scanner;

ReplyDeletepublic class Solution {

public static void main(String[] args) {

Scanner s = new Scanner(System.in);

int S = s.nextInt(); // 0

int E = s.nextInt(); //100

int dif = s.nextInt(); //10

while(S<=E){

int cel = ((S-32)*5)/9;

System.out.println(S+"\t"+cel);

S=S+dif;

}

}

}

I THINK THIS WILL SURELY HELP YOU.

thx

DeleteThis should do the job:

ReplyDelete#include

using namespace std;

int main()

{

int S,E,W,i,cd;

cin>>S;

cin>>E;

cin>>W;

while (((S>=0)&&(S<=80)) && ((E>=S)&&(S<=900)) && ((W>=0)&&(W<=40)))

{

while(S<=E)

{

for(i=S;i<=E;i+=W)

{

cd=(i-32)/1.8;

cout<<i<<"\t"<<cd<<"\n";

S+=W;

}

}

}

}

nice

ReplyDelete