# FACEDIR - Editorial

Setter: Utkarsh Gupta
Tester: Samarth Gupta
Editorialist: Taranpreet Singh

Cakewalk

None

# PROBLEM

The chef is standing facing North. For the next X second, he rotates exactly 90 degrees in the clockwise direction. Find the direction of Chef after X seconds where 1 \leq X \leq 100

Note: There are only 4 directions: North, East, South, West (in clockwise order).

# QUICK EXPLANATION

• After every 4 rotation, the Chef faces the same direction as before. So let’s group all rotations in buckets of 4. The last bucket may be unfilled.
• All buckets containing 4 rotations do not change the direction. We only care about the last bucket, which may have less than 4 rotations.
• We can simulate the last three rotations or write cases to identify the final direction.

# EXPLANATION

Since the constraints of this problem are small, a simulation-based solution is acceptable along the following lines.

dir = 'N'
repeat X times:
if(dir == 'N')dir = 'E'
else if(dir == 'E')dir = 'S'
else if(dir == 'S')dir = 'W'
else if(dir == 'W')dir = 'N'


The final value stored in dir represents the final direction. This solution takes O(X) time.

We can also solve it in O(1) as well, by noticing that rotating 4 times is equivalent to not rotating at all.

So we can say that doing X rotation gets same direction as doing X - 4 rotations, X-8 rotations … and (X - 4*C) for some C where X-4*C \geq 0. We can choose C = \lfloor X/4 \rfloor, which makes X-4*C = X \bmod 4.

Hence, we only need to perform up to 3 rotations, as we can replace X with X \bmod 4. We can use the above simulation or write cases to find out the final direction.

# TIME COMPLEXITY

The time complexity is O(X) or O(1) per test case.

# SOLUTIONS

Setter's Solution
//Utkarsh.25dec
#include <bits/stdc++.h>
#include <chrono>
#include <random>
#define ll long long int
#define ull unsigned long long int
#define pb push_back
#define mp make_pair
#define mod 1000000007
#define rep(i,n) for(ll i=0;i<n;i++)
#define loop(i,a,b) for(ll i=a;i<=b;i++)
#define vi vector <int>
#define vs vector <string>
#define vc vector <char>
#define vl vector <ll>
#define all(c) (c).begin(),(c).end()
#define max3(a,b,c) max(max(a,b),c)
#define min3(a,b,c) min(min(a,b),c)
#define deb(x) cerr<<#x<<' '<<'='<<' '<<x<<'\n'
using namespace std;
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
#define ordered_set tree<int, null_type,less<int>, rb_tree_tag,tree_order_statistics_node_update>
// ordered_set s ; s.order_of_key(val)  no. of elements strictly less than val
// s.find_by_order(i)  itertor to ith element (0 indexed)
typedef vector<vector<ll>> matrix;
ll power(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
ll modInverse(ll a){return power(a,mod-2);}
const int N=500023;
bool vis[N];
void solve()
{
ll n;
cin>>n;
n%=4;
if(n==0)
cout<<"North\n";
else if(n==1)
cout<<"East\n";
else if(n==2)
cout<<"South\n";
else if(n==3)
cout<<"West\n";
}
int main()
{
#ifndef ONLINE_JUDGE
freopen("input.txt", "r", stdin);
freopen("output.txt", "w", stdout);
#endif
//ios_base::sync_with_stdio(false);
//cin.tie(NULL);
int T=1;
cin>>T;
int t=0;
while(t++<T)
{
//cout<<"Case #"<<t<<":"<<' ';
solve();
//cout<<'\n';
}
cerr << "Time : " << 1000 * ((double)clock()) / (double)CLOCKS_PER_SEC << "ms\n";
}

Tester's Solution
#include <bits/stdc++.h>
using namespace std;

long long readInt(long long l, long long r, char endd) {
long long x=0;
int cnt=0;
int fi=-1;
bool is_neg=false;
while(true) {
char g=getchar();
if(g=='-') {
assert(fi==-1);
is_neg=true;
continue;
}
if('0'<=g&&g<='9') {
x*=10;
x+=g-'0';
if(cnt==0) {
fi=g-'0';
}
cnt++;
assert(fi!=0 || cnt==1);
assert(fi!=0 || is_neg==false);

assert(!(cnt>19 || ( cnt==19 && fi>1) ));
} else if(g==endd) {
if(is_neg) {
x=-x;
}
assert(l<=x&&x<=r);
return x;
} else {
assert(false);
}
}
}
string readString(int l, int r, char endd) {
string ret="";
int cnt=0;
while(true) {
char g=getchar();
assert(g!=-1);
if(g==endd) {
break;
}
cnt++;
ret+=g;
}
assert(l<=cnt&&cnt<=r);
return ret;
}
long long readIntSp(long long l, long long r) {
}
long long readIntLn(long long l, long long r) {
}
string readStringLn(int l, int r) {
}
string readStringSp(int l, int r) {
}

assert(getchar()==EOF);
}

int main() {
// your code goes here
ios_base::sync_with_stdio(false);
cin.tie(0);
int t;
t = readIntLn(1, 100);
int sum = 0;
while(t--){
int n;
n = readIntLn(1, 1000);
if(n%4 == 0)
cout << "North\n";
else if(n%4 == 1)
cout << "East\n";
else if(n%4 == 2)
cout << "South\n";
else
cout << "West\n";
}
return 0;
}

Editorialist's Solution
import java.util.*;
import java.io.*;
class FACEDIR{
//SOLUTION BEGIN
void pre() throws Exception{}
void solve(int TC) throws Exception{
String[] D = new String[]{"North", "East", "South", "West"};
int dir = ni()%4;
pn(D[dir]);
}
//SOLUTION END
void hold(boolean b)throws Exception{if(!b)throw new Exception("Hold right there, Sparky!");}
static boolean multipleTC = true;
void run() throws Exception{
in = new FastReader();
out = new PrintWriter(System.out);
//Solution Credits: Taranpreet Singh
int T = (multipleTC)?ni():1;
pre();for(int t = 1; t<= T; t++)solve(t);
out.flush();
out.close();
}
public static void main(String[] args) throws Exception{
new FACEDIR().run();
}
int bit(long n){return (n==0)?0:(1+bit(n&(n-1)));}
void p(Object o){out.print(o);}
void pn(Object o){out.println(o);}
void pni(Object o){out.println(o);out.flush();}
String n()throws Exception{return in.next();}
String nln()throws Exception{return in.nextLine();}
int ni()throws Exception{return Integer.parseInt(in.next());}
long nl()throws Exception{return Long.parseLong(in.next());}
double nd()throws Exception{return Double.parseDouble(in.next());}

StringTokenizer st;
}

public FastReader(String s) throws Exception{
}

String next() throws Exception{
while (st == null || !st.hasMoreElements()){
try{
st = new StringTokenizer(br.readLine());
}catch (IOException  e){
throw new Exception(e.toString());
}
}
return st.nextToken();
}

String nextLine() throws Exception{
String str = "";
try{
}catch (IOException e){
throw new Exception(e.toString());
}
return str;
}
}
}


Feel free to share your approach. Suggestions are welcomed as always.

import java.util.*;
public class evenodd
{
public static void main(String args[])
{
Scanner in=new Scanner(System.in);
int t,i;
System.out.println(“Enter test cases”);
t=in.nextInt();
int a[]=new int[t];
System.out.println(“Enter second”);
for(i=0;i<t;i++)
{
a[i]=in.nextInt();
}
for(i=0;i<t;i++)
{
if(a[i]%4==0)
System.out.println(“North”);
if(a[i]%4==1)
System.out.println(“East”);
if(a[i]%4==2)
System.out.println(“South”);
if(a[i]%4==3)
System.out.println(“West”);
}
}
}

please run this code.Somesort of error is coming .I’m unable to deal with it. Please help anyone