In this HackerEarth Micro and Sweet Distribution problem solution It's sweet distribution day in Micro's school. He's very happy. All the students in Micro's class are sitting on chairs which are arranged in a matrix of size N x M i.e. there are N rows of chairs numbered from 1 to N and in each row there are M chairs numbered from 1 to M. Micro is sitting at coordinate (Dx, Dy) (Dthy chair of Dthx row). Teacher gives the box to a student sitting in one of the four corners: (1,1), (1,M), (N,1) or (N,M). Students have to take one sweet from the box and pass the box to the next student (student sitting to left, right, front or back). For a student sitting at coordinate , he'll follow the following priority order:

If there is a student in the Right who has not received sweet, then pass it right (x, y+1).
If there is a student in the Left who has not received sweet, then pass it left (x, y-1).
If there is a student in the Front who has not received sweet, then pass it front (x-1,y).
If there is a student in the Back who has not received sweet, then pass it back (x+1,y).
Shout Over, meaning that all students have received sweets.
Now, Micro is curious to find out the direction in which he'll have to pass the sweet box. Since there are a lot of students in Micro's class, it will take long for the box to reach him, and you know Micro, he just can't wait. So he asks you to find out the direction in which he'll have to pass the box, or will he have to shout Over.

## HackerEarth Micro and Sweet Distribution problem solution.

`#include<bits/stdc++.h>using namespace std;int main(){    int t;cin>>t;    while(t--){        int n, m;cin>>n>>m;        int sx, sy;cin>>sx>>sy;        int x, y;cin>>x>>y;        if(sx == 1 and sy == 1){            if(x&1){                if(y == m){                    if(x == n)cout<<"Over";                    else cout<<"Back";                }                else cout<<"Right";            }            else{                if(y == 1){                    if(x == n)cout<<"Over";                    else cout<<"Back";                }                else cout<<"Left";            }        }        else if(sx == 1 and sy == m){            if(x&1){                if(y == 1){                    if(x == n)cout<<"Over";                    else cout<<"Back";                }                else cout<<"Left";            }            else{                if(y == m){                    if(x == n)cout<<"Over";                    else cout<<"Back";                }                else cout<<"Right";            }           }        else if(sx == n and sy == 1){            int dx = n-x+1;            if(dx&1){                if(y == m){                    if(x == 1)cout<<"Over";                    else cout<<"Front";                }                else cout<<"Right";            }            else{                if(y == 1){                    if(x == 1)cout<<"Over";                    else cout<<"Front";                }                else cout<<"Left";            }        }        else{            int dx = n-x+1;            if(dx&1){                if(y == 1){                    if(x == 1)cout<<"Over";                    else cout<<"Front";                }                else cout<<"Left";            }            else{                if(y == m){                    if(x == 1)cout<<"Over";                    else cout<<"Front";                }                else cout<<"Right";            }        }        cout<<endl;    }    return 0;}`

### Second solution

`#include<bits/stdc++.h>#include<iostream>using namespace std;#define pin(s) {cout<<(s)<<endl;continue;}int main(){    int T,N,M,sx,sy,dx,dy;    cin>>T;    while(T--){        cin>>N>>M>>sx>>sy>>dx>>dy;        int ox = N+1-sx, oy = N%2==0?sy:M+1-sy;        if(ox==dx and oy==dy)            pin("Over")        if((dx-sx)%2==0 and dy==M+1-sy or (dx-sx)%2!=0 and dy==sy)            pin(sx==1?"Back":"Front")        if((dx-sx)%2==0)pin(sy==1?"Right":"Left")        else pin(sy==M?"Right":"Left")    }}`