HackerEarth Rahuls Logo problem solution

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

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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.

#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;
}

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