Given three integers N, A, and B. A person is standing at 0-th coordinate and moves A steps to the right in the first step, B steps to the left at the second step, and so on. The task is to find out at which coordinate he will be after N steps.
Examples:
Input: N = 3, A = 5 and B = 2
Output: 8
5 to the right, 2 to the left and 5 to the right, hence the person will end at 8.
Input: N = 5, A = 1 and B = 10
Output: -17
Approach: Since the person takes the odd number step to right and even the number of steps to the left, we have to find out the number difference in steps in either direction. Hence, the formula obtained will be thus:
[((n+1)/2)*a - (n/2)*b]
Below is the implementation of the above approach:
C++
// C++ program to find the last coordinate// where it ends his journey#include <bits/stdc++.h>using namespace std;// Function to return the last destinationint lastCoordinate(int n, int a, int b){ return ((n + 1) / 2) * a - (n / 2) * b;}// Driver Codeint main(){ int n = 3, a = 5, b = 2; cout << lastCoordinate(n, a, b); return 0;} |
Java
// Java program to find the last coordinate// where it ends his journeyimport java.util.*;class solution{// Function to return the last destinationstatic int lastCoordinate(int n, int a, int b){ return ((n + 1) / 2) * a - (n / 2) * b;}// Driver Codepublic static void main(String args[]){ int n = 3, a = 5, b = 2; System.out.println(lastCoordinate(n, a, b));}} |
Python
# Python3 program to find the # last coordinate where it# ends his journey # Function to return the# last destination def lastCoordinate(n, a, b): return (((n + 1) // 2) * a - (n // 2) * b)# Driver Code n = 3a = 5b = 2print(lastCoordinate(n, a, b))# This code is contributed# by Sanjit_Prasad |
C#
// C# program to find the last coordinate // where it ends his journey using System;class GFG{// Function to return the last destination public static int lastCoordinate(int n, int a, int b){ return ((n + 1) / 2) * a - (n / 2) * b;}// Driver Code public static void Main(string[] args){ int n = 3, a = 5, b = 2; Console.WriteLine(lastCoordinate(n, a, b));}}// This code is contributed by Shrikant13 |
PHP
<?php// PHP program to find the last coordinate// where it ends his journey// Function to return the last destinationfunction lastCoordinate($n, $a, $b){ return (($n + 1) / 2) * $a - (int)($n / 2) * $b;}// Driver Code$n = 3; $a = 5; $b = 2;echo lastCoordinate($n, $a, $b);// This code is contributed by inder_verma..?> |
Javascript
<script>// Javascript program to find// the last coordinate// where it ends his journey// Function to return the last destinationfunction lastCoordinate(n, a, b){ return (parseInt(n + 1) / 2) * a - parseInt(n / 2) * b;}// Driver Code let n = 3, a = 5, b = 2; document.write(lastCoordinate(n, a, b));</script> |
Output:
8
Time Complexity: O(1)
Auxiliary Space: O(1)
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