In this HackerEarth Rahul's Logo problem solution Rahul has set upon the quest for a new logo of his company. He has created the following continuous logo:

    /\
   /  \
  / /\ \
 / /  \ \
/ / /\ \ \
  \ \ \/ / /
   \ \  / /
    \ \/ /
     \  /
      \/
However, his sister, Rashi, likes the following discontinuous design more

   /\
  /  \
 / /\ \
/ /  \ \
  \ \  / /
   \ \/ /
    \  /
     \/
The size of a logo is the longest continuous streak of the same characters on an arm. So, the size of 1st logo is 5 while that of 2nd one is 4. Now, he wants to paint both of these logos on a canvas.

Given an integer N, write a program that outputs these logos for sizes N and N + 1, one below the other. Please note that Rahul's logo is only valid for odd sizes and Rashi's logo is only valid for even values.



HackerEarth Rahul's Logo problem solution


HackerEarth Rahul's Logo problem solution.

#include <bits/stdc++.h>
using namespace std;

#define MOD                     1000000007
#define pb(x)                   push_back(x)
#define mp(x,y)                 make_pair(x,y)
#define FF                      first
#define SS                      second
#define s(n)                    scanf("%d",&n)
#define sl(n)                   scanf("%lld",&n)
#define sf(n)                   scanf("%lf",&n)
#define ss(n)                   scanf("%s",n)
#define sc(n)                   {char temp[4]; ss(temp); n=temp[0];}
#define INF                     (int)1e9
#define LINF                    (long long)1e18
#define EPS                     1e-9
#define maX(a,b)                ((a)>(b)?(a):(b))
#define miN(a,b)                ((a)<(b)?(a):(b))
#define abS(x)                  ((x)<0?-(x):(x))

typedef long long ll;
typedef unsigned long long LL;
typedef pair<int,int> PII;
typedef pair<ll,ll> Pll;
typedef pair<int,PII> TRI;
typedef vector<int> VI;
typedef vector<ll> VL;
typedef vector<ll> vl;
typedef vector<PII> VII;
typedef vector<TRI> VT;

int n1, n2;
char a[1000][1000];

void precompute() {
}
void read() {
  s(n1);
    n2 = n1 + 1;
}
void preprocess() {

}

void make_diamond(int i, int j, int num) {
  if(num <= 0) return;
  int front = j, back = j + 1;
  int d[] = {-1, 1}, ch[] = {'/','\\'};
  int cur = 0;
  int ff = 0, bb = 1, off = 0;    
  while(front < back) {
    a[i + cur][front + off] = ch[ff];
    a[i + cur][back + off] = ch[bb];        
    if(cur == num - 1) {
      swap(ff, bb);
      cur++;
      off = 2;
      a[i + cur][front + off] = ch[ff];
      a[i + cur][back + off] = ch[bb]; 
    }
    front += d[ff];
    back += d[bb];
    cur++;
  }
  make_diamond(i + 2, j, num - 2);
}
void print_logo(int n) {
    memset(a, 0, sizeof a);
  int N = 2 * n;
    make_diamond(0, n - 1, n);
    for (int i = 0; i < N; ++i) {
      int p1 = 0, p2 = 0;
      while(1) {      
        while(p2 < N + 2 and a[i][p2] == 0) p2++;
            if(p2 == N + 2) break;
        while(p1 < p2) a[i][p1++] = ' ';
        p1 = ++p2;
      }
    }
    for (int i = 0; i < N; ++i) {
      puts(a[i]);
    }
}

void solve() {
    print_logo(n1);
    print_logo(n2);
}
int main() {
  precompute();
    read();
    preprocess();
    solve();
    return 0;
}


Second solution

#include<bits/stdc++.h>
#define PB push_back
#define MP make_pair
#define F first
#define S second
#define SZ(a) (int)(a.size())
#define SET(a,b) memset(a,b,sizeof(a))
#define LET(x,a) __typeof(a) x(a)
#define TR(v,it) for( LET(it,v.begin()) ; it != v.end() ; it++)
#define repi(i,n) for(int i=0; i<(int)n;i++)
#define si(n) scanf("%d",&n)
#define sll(n) scanf("%lld",&n)
#define sortv(a) sort(a.begin(),a.end())
#define all(a) a.begin(),a.end()
#define DRT()  int t; cin>>t; while(t--)
#define TRACE
using namespace std;

#ifdef TRACE
#define trace1(x)                cerr << #x << ": " << x << endl;
#define trace2(x, y)             cerr << #x << ": " << x << " | " << #y << ": " << y << endl;
#define trace3(x, y, z)          cerr << #x << ": " << x << " | " << #y << ": " << y << " | " << #z << ": " << z << endl;
#define trace4(a, b, c, d)       cerr << #a << ": " << a << " | " << #b << ": " << b << " | " << #c << ": " << c << " | " << #d << ": " << d << endl;
#define trace5(a, b, c, d, e)    cerr << #a << ": " << a << " | " << #b << ": " << b << " | " << #c << ": " << c << " | " << #d << ": " << d << " | " << #e << ": " << e << endl;
#define trace6(a, b, c, d, e, f) cerr << #a << ": " << a << " | " << #b << ": " << b << " | " << #c << ": " << c << " | " << #d << ": " << d << " | " << #e << ": " << e << " | " << #f << ": " << f << endl;

#else

#define trace1(x)
#define trace2(x, y)
#define trace3(x, y, z)
#define trace4(a, b, c, d)
#define trace5(a, b, c, d, e)
#define trace6(a, b, c, d, e, f)

#endif


typedef long long LL;
typedef pair<int,int> PII;
typedef vector<int> VI;
typedef vector< PII > VPII;


void g(int n)
{
    for(int i=0; i<n;i++)
    {
        for(int j=0; j<2+i; j++)cout<<" ";
        for(int j=0; j<n-i; j++)
            if(j%2==0)cout<<"\\"; 
            else cout<<" ";
        for(int j=n-i; j>0; j--)
            if(j%2)cout<<"/"; 
            else cout<<" ";
        cout<<endl;
    }
}
void f(int n)
{
    for(int i=0; i<n;i++)
    {
        for(int j=1; j < n-i; j++)cout<<" ";
        for(int j=0; j<i+1; j++)
            if(j%2==0)cout<<"/"; 
            else cout<<" ";
        for(int j=0; j<i+1; j++)
            if((i-j)%2==0)cout<<"\\"; 
            else cout<<" ";
        cout<<endl;
    }
}

int main()
{
        int n; cin>>n; 
        f(n); g(n);
        n++; f(n); g(n);
    return 0;
}