Arithmetic Progression - HackerEarth Problem Solution

You will be given three numbers A,B,C .You can perform the following operation on these numbers any number of times.You can take any integer from A, B, C and you can add or substract 1 from it.

Each operation cost 1sec of time(say). Now you have to determine the minimum time required to change those numbers into an Arithmetic Progression. 

i.e B-A=C-B

Input :

First line of input contains T denoting number of test cases.

Next T lines contains space seperated integers A,B,C

Output :

For each test case, print a single line containing one integer — the minimum time require to change A,B,C into an arithmetic progression.

Arithmetic Progression - HackerEarth Problem Solution

In C - 

#include<stdio.h> 

#include<math.h>

int main() 

{ 

long long int t,a,b,c,count; 

int i;

scanf("%lld",&t); 

for(i=0;i<t;i++) 

{ 

scanf("%lld %lld %lld",&a,&b,&c); 

count=abs((2*b-(a+c))); 

if(count%2==0) 

printf("%d\n",count/2); 

else 

printf("%d\n",count/2+1);

}

}

Arithmetic Progression - HackerEarth Problem Solution

In C++ - 

#include<bits/stdc++.h>

using namespace std;

int main() {

    ios::sync_with_stdio(0);

    cin.tie(0);

    int t;

    cin>> t;

    while(t--){

        int a,b,c;

        cin>>a>>b>>c;

        int k, count=0;

        k= abs((2*b)-(a+c));

        (k%2==0)?count=k/2:count=(k+1)/2;

        cout<<count<<"\n";

    }

    

}

Arithmetic Progression - HackerEarth Problem Solution

In Java - 

import java.io.BufferedReader;

import java.io.InputStreamReader;

class TestClass {

    public static void main(String args[] ) throws Exception {

        

        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

        int tests= Integer.parseInt(br.readLine());

        StringBuilder out = new StringBuilder();

        for(int x=0;x<tests;x++){

            String arr[]=br.readLine().split(" ");

            int a = Integer.parseInt(arr[0]);

            int b=  Integer.parseInt(arr[1]);

            int c=  Integer.parseInt(arr[2]);

         

            int diff = (Math.abs((c-b)-(b-a))+1)/2;

          

            out.append(diff+"\n");

        }

        System.out.println(out);

    }

}

Arithmetic Progression - HackerEarth Problem Solution

In Python - 

test_case=int(input())

output=""

i=0

while i<test_case:

    i+=1

    lst=[]

    lst=input().split(' ')

    mean=(float(lst[0])+float(lst[2]))/2

    if mean == float(lst[1]):

        output+="0\n"

    elif mean >= float(lst[1]):

        output+=str(int(mean)-int(lst[1])+int((mean-int(mean))*2))+"\n"

    else:

        output+=str(int(lst[1])-int(mean))+"\n"

print(output)