Kangaroo

There are two kangaroos on an x-axis ready to jump in the positive direction (i.e, toward positive infinity). The first kangaroo starts at location  and moves at a rate of  meters per jump. The second kangaroo starts at location  and moves at a rate of  meters per jump. Given the starting locations and movement rates for each kangaroo, can you determine if they'll ever land at the same location at the same time?

Input Format

A single line of four space-separated integers denoting the respective values of , , , and .

Constraints

• 0<= x1 < x2 <= 10000
• 1 <= x1 <= 10000
• 1<= v2 <= 10000

Output Format

Print YES if they can land on the same location at the same time; otherwise, print NO.

MY EXPLANATION: 

This is a simple high school maths problem like 2 trains start at different location,given their velocity can you find whether they will ever meet or not?

Let's name the kangaroo's as K1 and K2 and given K2 starts ahead of K1 (i.e. x2 > x1).

Now, they will meet only when distance traveled by K1 is equal to the sum of distance traveled by K2 and their initial separation(i.e. (x1-x2) + distance(K2) ). This is visualized below.

let say they meet after time T then the condition will be

            v1*T = (x2 - x1) + v2*T

In other word, they will meet if

T = ( x2 - x1 ) % ( v1 - v2 )  is equal to 0 ( zero)

Now try to write your own code keeping these things in mind.

If you don't succeed than take a look at my solution.

#include <iostream>

using namespace std;

int main(){   

int x1, x2, v1, v2; /* Initializing variables */ 

cin >> x1 >> v1 >> x2 >> v2; /* assigining values to the variables  */ 

if( v1>v2  &&  (x2-x1)%(v1-v2)==0 )cout<<"YES"; /* if this is the case than they will meet , so print YES */ 

else cout<<"NO"; /* In any other case they wont meet */   

return 0;

} 

Question Source: hackerrank.com