**Number Line Jumps**

### Problem Statement :

You are choreographing a circus show with various animals. For one act, you are given two kangaroos on a number line 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. You have to figure out a way to get both kangaroos at the same location at the same time as part of the show. If it is possible, return YES, otherwise return NO. For example, kangaroo 1 starts at x1=2 with a jump distance v1=1 and kangaroo 2 starts at x2=1 with a jump distance of v2=2. After one jump, they are both at x=3, (x1+v1 = 2+1, x2+v2 = 1+2 ), so our answer is YES. Function Description Complete the function kangaroo in the editor below. It should return YES if they reach the same position at the same time, or NO if they don't. kangaroo has the following parameter(s): x1, v1: integers, starting position and jump distance for kangaroo 1 x2, v2: integers, starting position and jump distance for kangaroo 2 Input Format A single line of four space-separated integers denoting the respective values of x1, v1, x2, and v2. Constraints 0 <= x1, x2 <= 10000 1 <= v1 <= 10000 1 <= v2 <= 10000 Output Format Print YES if they can land on the same location at the same time; otherwise, print NO. Note: The two kangaroos must land at the same location after making the same number of jumps.

### Solution :

` ````
Solution in C :
c :
#include <math.h>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <assert.h>
#include <limits.h>
#include <stdbool.h>
int main(){
int x1;
int v1;
int x2;
int v2;
scanf("%d %d %d %d",&x1,&v1,&x2,&v2);
if(x2>x1){
if(v2<v1){
if((x2-x1)%(v1-v2)==0)
printf("YES");
else
printf("NO");
}
else
printf("NO");
}
else{
if(v1>v2){
if((x1-x2)%(v2-v1)==0)
printf("YES");
else
printf("NO");
}
else if(x1==x2&&v1==v2)
printf("YES");
else
printf("NO");
}
return 0;
}
C++ :
#include <map>
#include <set>
#include <list>
#include <cmath>
#include <ctime>
#include <deque>
#include <queue>
#include <stack>
#include <string>
#include <bitset>
#include <cstdio>
#include <limits>
#include <vector>
#include <climits>
#include <cstring>
#include <cstdlib>
#include <fstream>
#include <numeric>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <unordered_map>
using namespace std;
int main(){
int x1;
int v1;
int x2;
int v2;
cin >> x1 >> v1 >> x2 >> v2;
if ((v1 <= v2) || ((x2 - x1) % (v2 - v1))) {
puts("NO");
} else {
puts("YES");
}
return 0;
}
Java :
import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;
public class Solution {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int x1 = in.nextInt();
int v1 = in.nextInt();
int x2 = in.nextInt();
int v2 = in.nextInt();
if(x1>x2)
{
if(v1<v2)
{
for(;;)
{
x1=x1+v1;
x2=x2+v2;
if (x1==x2)
{
System.out.println("YES");
break;
}
else
{
if (x2>x1)
{
System.out.println("NO");
break;
}
}
}
}
else
System.out.println("NO");
}
else if(x1<x2)
{
if(v1>v2)
{
for(;;)
{
x1=x1+v1;
x2=x2+v2;
if (x1==x2)
{
System.out.println("YES");
break;
}
else
{
if (x1>x2)
{
System.out.println("NO");
break;
}
}
}
}
else
System.out.println("NO");
}
else
{
if(v1==v2)
System.out.println("YES");
else
System.out.println("NO");
}
}
}
python3 :
a, b, c, d = map(int, input().split(' '))
if a-c==0:
print("YES")
quit()
if d-b==0:
print("NO")
quit()
if ((a-c)%(d-b)==0 and (a-c)//(d-b)>=0):
print("YES")
quit()
print("NO")
```

## View More Similar Problems

## Kundu and Tree

Kundu is true tree lover. Tree is a connected graph having N vertices and N-1 edges. Today when he got a tree, he colored each edge with one of either red(r) or black(b) color. He is interested in knowing how many triplets(a,b,c) of vertices are there , such that, there is atleast one edge having red color on all the three paths i.e. from vertex a to b, vertex b to c and vertex c to a . Note that

View Solution →## Super Maximum Cost Queries

Victoria has a tree, T , consisting of N nodes numbered from 1 to N. Each edge from node Ui to Vi in tree T has an integer weight, Wi. Let's define the cost, C, of a path from some node X to some other node Y as the maximum weight ( W ) for any edge in the unique path from node X to Y node . Victoria wants your help processing Q queries on tree T, where each query contains 2 integers, L and

View Solution →## Contacts

We're going to make our own Contacts application! The application must perform two types of operations: 1 . add name, where name is a string denoting a contact name. This must store name as a new contact in the application. find partial, where partial is a string denoting a partial name to search the application for. It must count the number of contacts starting partial with and print the co

View Solution →## No Prefix Set

There is a given list of strings where each string contains only lowercase letters from a - j, inclusive. The set of strings is said to be a GOOD SET if no string is a prefix of another string. In this case, print GOOD SET. Otherwise, print BAD SET on the first line followed by the string being checked. Note If two strings are identical, they are prefixes of each other. Function Descriptio

View Solution →## Cube Summation

You are given a 3-D Matrix in which each block contains 0 initially. The first block is defined by the coordinate (1,1,1) and the last block is defined by the coordinate (N,N,N). There are two types of queries. UPDATE x y z W updates the value of block (x,y,z) to W. QUERY x1 y1 z1 x2 y2 z2 calculates the sum of the value of blocks whose x coordinate is between x1 and x2 (inclusive), y coor

View Solution →## Direct Connections

Enter-View ( EV ) is a linear, street-like country. By linear, we mean all the cities of the country are placed on a single straight line - the x -axis. Thus every city's position can be defined by a single coordinate, xi, the distance from the left borderline of the country. You can treat all cities as single points. Unfortunately, the dictator of telecommunication of EV (Mr. S. Treat Jr.) do

View Solution →