Hackerrank - HourRank 15 - 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.

Solution:

1. #include <bits/stdc++.h>
2. #define pb push_back
3. using namespace std;
4. typedef long long LL;
5. const int S = 100003;
6. const int MOD = 1e9+7;
7. const double pi = 2 * acos (0.0);
8.  
9. int _Int(){int x; scanf("%d", &x); return x;}
10. LL _LLi(){LL x; scanf("%lld", &x); return x;}
11. void pruts(){puts("-1");exit(EXIT_SUCCESS);}
12.  
13. int dirX[] = { 1, 0, -1, 0, 1, -1, 1, -1 };
14. int dirY[] = { 0, 1, 0, -1, 1, -1, -1, 1 };
15. int rX[] = { 1, 1, 2, 2, -1, -1, -2, -2 };
16. int rY[] = { 2, -2, 1, -1, 2, -2, 1, -1 };
17.  
18. template < class T > T tri_Area( T x1, T y1, T x2, T y2, T x3, T y3 ){ return abs( x1*( y2-y3 ) - y1*( x2-x3 ) + ( x2*y3-x3*y2 ) );};
19. template < class T > T Distance( T x1, T y1, T x2, T y2 ){ return sqrt( ( x1-x2 ) * ( x1-x2 ) + ( y1-y2 ) * ( y1-y2 ) ); };
20. template < class T > T bigMod( T n, T p, T m ){ if( p == 0 )return 1; if( p&1 )return ( n*bigMod( n, p-1, m ) )%m; T x = bigMod( n, p/2, m ); return ( x*x )%m; };
21.  
22. int sset(int N, int pos){return N=N|(1<<pos);}
23. bool check(int N, int pos){return (bool)(N&(1<<pos));}
24. int reset(int N, int pos){return N=N&~(1<<pos);}
25. /*******************###########################################********************
26. ********************##   MD. YA-SEEN ARAFAT(ThunderStroke)   ##********************
27. ********************##    CSE, University of Asia Pacific    ##********************
28. ********************###########################################********************/
29.  
30. void Love(){
31.     int q = _Int();
32.     while( q-- ){
33.         int a = _Int();
34.         int b = _Int();
35.         int c = _Int();
36.         int A = 0, B = 0, C = 0;
37.         if( abs( a-c ) > abs( b-c ) )B = 1;
38.         else if( abs( a-c ) < abs( b-c ) )A = 1;
39.         else C = 1;
40.         if( A )puts( "Cat A" );
41.         if( B )puts( "Cat B" );
42.         if( C )puts( "Mouse C" );
43.     }
44. }
45.  
46. int main(){
47.     #ifndef ONLINE_JUDGE
48. //        freopen("in.txt", "r", stdin);
49. //        freopen("n.txt", "w", stdout);
50.     #endif
51.     Love();
52.     return 0;
53. }