## Saturday, December 3, 2016

### Cats and a Mouse

Two cats named  and  are standing at integral points on the x-axis. Cat  is standing at point  and cat  is standing at point . Both cats run at the same speed, and they want to catch a mouse named  that's hiding at integral point  on the x-axis. Can you determine who will catch the mouse?
You are given  queries in the form of , and . For each query, print the appropriate answer on a new line:
• If cat  catches the mouse first, print Cat A.
• If cat  catches the mouse first, print Cat B.
• If both cats reach the mouse at the same time, print Mouse C as the two cats fight and mouse escapes.
Input Format
The first line contains a single integer, , denoting the number of queries.
Each of the  subsequent lines contains three space-separated integers describing the respective values of  (cat 's location),  (cat 's location), and  (mouse 's location).
Constraints
Output Format
On a new line for each query, print Cat A if cat  catches the mouse first, Cat B if cat  catches the mouse first, or Mouse C if the mouse escapes.
Sample Input 0
3
1 2 3
1 3 2
2 1 3

Sample Output 0
Cat B
Mouse C
Cat A

Explanation 0
Query 0: The positions of the cats and mouse are shown below:
Cat  will catch the mouse first, so we print Cat B on a new line.
Query 1: In this query, cats  and  reach mouse  at the exact same time:
Because the mouse escapes, we print Mouse C on a new line.
Note: WRITE YOUR PROGRAM IN COMMENTS SECTION

***********
PROGRAM
***********
#include <math.h>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <assert.h>
#include <limits.h>
#include <stdbool.h>

int main(){
int q;
scanf("%d",&q);
for(int a0 = 0; a0 < q; a0++){
int x;
int y;
int z;
scanf("%d %d %d",&x,&y,&z);
int yz,xz;
yz=y-z;
if(yz<0)
yz=(-1)*yz;
xz=x-z;
if(xz<0)
xz=(-1)*xz;
if(xz<yz)
printf("Cat A\n");
else if(xz>yz)
printf("Cat B\n");
else
printf("Mouse C\n");
}
return 0;
}