- Prepare
- Algorithms
- Implementation
- Number Line Jumps

# Number Line Jumps

# Number Line Jumps

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 and moves at a rate of meters per jump.
- The second kangaroo starts at location and moves at a rate of 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`

.

**Example**

After one jump, they are both at , (, ), so the answer is `YES`

.

**Function Description**

Complete the function *kangaroo* in the editor below.

kangaroo has the following parameter(s):

*int x1, int v1*: starting position and jump distance for kangaroo 1*int x2, int v2*: starting position and jump distance for kangaroo 2

**Returns**

*string:*either`YES`

or`NO`

**Input Format**

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

**Constraints**

**Sample Input 0**

```
0 3 4 2
```

**Sample Output 0**

```
YES
```

**Explanation 0**

The two kangaroos jump through the following sequence of locations:

From the image, it is clear that the kangaroos meet at the same location (number on the number line) after same number of jumps ( jumps), and we print `YES`

.

**Sample Input 1**

```
0 2 5 3
```

**Sample Output 1**

```
NO
```

**Explanation 1**

The second kangaroo has a starting location that is ahead (further to the right) of the first kangaroo's starting location (i.e., ). Because the second kangaroo moves at a faster rate (meaning ) *and* is already ahead of the first kangaroo, the first kangaroo will never be able to catch up. Thus, we print *NO*.