Codeforces Round #259 (Div. 2) A. Little Pony and Crystal Mine-优快云博客

本文介绍了一个简单的编程挑战，即根据输入的奇数尺寸n，绘制一个包含钻石形状的水晶图案。文章提供了一段C++代码示例，展示了如何通过循环和条件判断来生成所需的图形。

Twilight Sparkle once got a crystal from the Crystal Mine. A crystal of size n (n is odd; n > 1) is an n × n matrix with a diamond inscribed into it.

You are given an odd integer 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

The only line contains an integer n (3 ≤ n ≤ 101; n is odd).

Output

Output a crystal of size 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***

题意：像sample那样输出一个图形。。

思路：这不是语言练习题么。。当时没时间了，没有仔细思考，胡乱写的，发现错位稍微改一下就对了。。

#include <iostream>  
#include <stdio.h>  
#include <cmath>  
#include <algorithm>  
#include <iomanip>  
#include <cstdlib>  
#include <string>  
#include <memory.h>  
#include <vector>  
#include <queue>  
#include <stack>  
#include <map>
#include <set>
#define ll long long
#define INF 1000000

using namespace std;


int main(){
	int n;
	while(cin>>n){
		for(int i=1;i<=n;i++){
			int tmp=i;
			if(i>n/2+1)tmp=n+1-i;
			for(int j=1;j<=n;j++){
				if(j>=n/2+1-(tmp-1)&&j<=n/2+1+(tmp-1)){
					printf("D");
				}else
					printf("*");
					
			}
			cout<<endl;
		}
	}
	return 0;
}