CF 454A(Little Pony and Crystal Mine-画图[水])

A. Little Pony and Crystal Mine

time limit per test

memory limit per test

input

output

n (n is odd; n > 1) is an n × n

n. You need to draw a crystal of size n. The diamond cells of the matrix should be represented by character "D". All other cells of the matrix should be represented by character "*". Look at the examples to understand what you need to draw.

Input

n (3 ≤ n ≤ 101; n

Output

n.

Sample test(s)

input

3

output

*D* DDD *D*

input

5

output

**D** *DDD* DDDDD *DDD* **D**

input

7

output

***D*** **DDD** *DDDDD* DDDDDDD *DDDDD* **DDD** ***D***

水题，直接画图

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<cctype>
#include<ctime>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Lson (x<<1)
#define Rson ((x<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (100000007)
long long mul(long long a,long long b){return (a*b)%F;}
long long add(long long a,long long b){return (a+b)%F;}
long long sub(long long a,long long b){return (a-b+(a-b)/F*F+F)%F;}
typedef long long ll;
int main()
{
//  freopen("Crystal Mine.in","r",stdin);
//  freopen(".out","w",stdout);
  int n;
  cin>>n;
  int t=(n-1)/2,p=(n+1)/2;
  For(i,n)
  {
    For(j,n)
      if (abs(p-i)+abs(p-j)<=t) printf("D");else printf("*");
    
    
    cout<<endl;
  }
  
  
  return 0;
}