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
YESorNO
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.
SOLUTION:
#include<stdio.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;
}
