ProblemThere 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 x1 and moves at a rate of v1 meters per jump. The second  kangaroo starts at location x2 and moves at a rate of v2 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?My Solution#include <iostream>using namespace std;int main(){    int x1, v1, x2, v2;    cin >> x1 >> v1 >> x2 >> v2;    // If one kangaroo is behind the other AND moving slower,     //    he/she will never catch up to the other one    if ((x1 < x2) && (v1 < v2)) cout << NO;    else if ((x2 < x1) && (v2 < v1)) cout << NO;    // Otherwise, move each kangaroo one jump at a time until     //     the one behind is no longer behind.     else {        if (x1 < x2) {            while (x1 < x2) {                x1 += v1;                x2 += v2;            }        } else {            while (x2 < x1) {                x1 += v1;                x2 += v2;            }        }        // Once he/she is no longer behind the other, check to see        //    if he/she is in the same position, or if he/she has passed        if (x1 == x2) cout << YES;        else cout << NO;    }    return 0;}I am wondering if there is a way to optimize my code for this particular HackerRank problem. I have gotten all of the correct answers, and none of them timeout so for the problem it's 'good enough,' but I am curious if there is a better way of doing it.Do I use too many conditionals, loops, etc? I have a bad habit of do everything in for loops, if statements, etc.
Determining if the kangaroos will land in the same position
c++;performance;programming challenge
Closed form solutionA better approach is to calculate the intercept point directly and then check whether or not it is an integer. The time of intercept is:$$t = \frac{x_2 - x_1}{v_1 - v_2}$$where both numerator and denominator are integers, but their ratio may not. We can also handle all special cases from the ratio.bool kangaroos_meet(int x1, int v1, int x2, int v2) {    int numerator = x2 - x1;    int denominator = v1 - v2;    if (denominator == 0)                // same velocity    {        return numerator == 0;           // they meet always or never    }    if (numerator % denominator != 0)    // intercept point not an integer    {        return false;    }    int t = numerator / denominator;     // calculate intercept point    return t >= 0;                       // intercept point lies in the past}