CSES Problem Set
Gray Code
A Gray code is a list of all bit strings of length , where any two successive strings differ in exactly one bit (i.e., their Hamming distance is one).
Your task is to create a Gray code for a given length .
Input
The only input line has an integer .
Output
Print lines that describe the Gray code. You can print any valid solution.
Constraints
Input:
Output:
Your task is to create a Gray code for a given length .
Input
The only input line has an integer .
Output
Print lines that describe the Gray code. You can print any valid solution.
Constraints
Input:
2
Output:
00
01
11
10
#include <bits/stdc++.h>
using namespace std;
#define int long long int
#define F first
#define S second
#define pb push_back
#define si set <int>
#define vi vector <int>
#define pii pair <int, int>
#define vpi vector <pii>
#define vpp vector <pair<int, pii>>
#define mii map <int, int>
#define mpi map <pii, int>
#define spi set <pii>
#define endl "\n"
#define sz(x) ((int) x.size())
#define all(p) p.begin(), p.end()
#define double long double
#define que_max priority_queue <int>
#define que_min priority_queue <int, vi, greater<int>>
#define bug(...) __f (#__VA_ARGS__, __VA_ARGS__)
#define print(a) for(auto x : a) cout << x << " "; cout << endl
#define print1(a) for(auto x : a) cout << x.F << " " << x.S << endl
#define print2(a,x,y) for(int i = x; i < y; i++) cout<< a[i]<< " "; cout << endl
inline int power(int a, int b)
{
int x = 1;
while (b)
{
if (b & 1) x *= a;
a *= a;
b >>= 1;
}
return x;
}
template <typename Arg1>
void __f (const char* name, Arg1&& arg1) { cout << name << " : " << arg1 << endl; }
template <typename Arg1, typename... Args>
void __f (const char* names, Arg1&& arg1, Args&&... args)
{
const char* comma = strchr (names + 1, ',');
cout.write (names, comma - names) << " : " << arg1 << " | "; __f (comma + 1, args...);
}
const int N = 200005;
void solve() {
int n;cin>>n;
vector<string>ans;
ans.push_back("0");
ans.push_back("1");
if(n==1){cout<<"0\n1\n";return;}
else{
while(n>1){
n--;
vector<string>l1=ans;
vector<string>l2=ans;
reverse(l2.begin(),l2.end());
int k=ans.size();
ans.clear();
for(int i=0; i<k; i++){
l1[i]='0'+l1[i];
ans.push_back(l1[i]);
}
for(int i=0; i<k; i++){
l2[i]='1'+l2[i];
ans.push_back(l2[i]);
}
}
for(auto c:ans)cout<<c<<endl;
}
}
int32_t main()
{
ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
clock_t z = clock();
int t = 1;
// cin >> t;
while (t--) solve();
return 0;
}
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