CF#259 (Div. 2) A.

Little Pony and Crystal Mine

time limit per test

1 second

memory limit per test

256 megabytes

题目链接：http://codeforces.com/contest/454/problem/A

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***

解题思路：

题意没细看，直接看样例了。样例给的很清楚，输入一个n，然后照那样输出就行。自从回了预处理之后，就欲罢不能，

这题也是用了下预处理。

先把二维矩阵g的元素都初始化为D，然后n行遍历。遍历分两个阶段，即 * 的数量增长阶段和 * 数量下降阶段。以减少

为例，经观察它最多可以减少到n / 2行。对于每行来说，分别对头和尾两次遍历。注意当k减少到0，开始增加回n / 2的阶段做

下标记，如果最后k增加到tmp + 1，则跳出循环······很好想，看代码吧

完整代码：

#include <functional>
#include <algorithm>
#include <iostream>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <numeric>
#include <cstring>
#include <climits>
#include <cassert>
#include <complex>
#include <cstdio>
#include <string>
#include <vector>
#include <bitset>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <list>
#include <set>
#include <map>

using namespace std;

#define REP(i, n) for (int i=0;i<n;++i)
#define FOR(i, a, b) for (int i=a;i<b;++i)
#define DWN(i, b, a) for (int i=b-1;i>=a;--i)
#define REP_1(i, n) for (LL i=1;i<=n;++i)
#define FOR_1(i, a, b) for (int i=a;i<=b;++i)
#define DWN_1(i, b, a) for (LL i=b;i>=a;--i)
#define REP_C(i, n) for (int n____=n,i=0;i<n____;++i)
#define FOR_C(i, a, b) for (int b____=b,i=a;i<b____;++i)
#define DWN_C(i, b, a) for (int a____=a,i=b-1;i>=a____;--i)
#define REP_N(i, n) for (i=0;i<n;++i)
#define FOR_N(i, a, b) for (i=a;i<b;++i)
#define DWN_N(i, b, a) for (i=b-1;i>=a;--i)
#define REP_1_C(i, n) for (int n____=n,i=1;i<=n____;++i)
#define FOR_1_C(i, a, b) for (int b____=b,i=a;i<=b____;++i)
#define DWN_1_C(i, b, a) for (int a____=a,i=b;i>=a____;--i)
#define REP_1_N(i, n) for (i=1;i<=n;++i)
#define FOR_1_N(i, a, b) for (i=a;i<=b;++i)
#define DWN_1_N(i, b, a) for (i=b;i>=a;--i)
#define REP_C_N(i, n) for (int n____=(i=0,n);i<n____;++i)
#define FOR_C_N(i, a, b) for (int b____=(i=0,b);i<b____;++i)
#define DWN_C_N(i, b, a) for (int a____=(i=b-1,a);i>=a____;--i)
#define REP_1_C_N(i, n) for (int n____=(i=1,n);i<=n____;++i)
#define FOR_1_C_N(i, a, b) for (int b____=(i=a,b);i<=b____;++i)
#define DWN_1_C_N(i, b, a) for (int a____=(i=b,a);i>=a____;--i)

#define ECH(it, A) for (__typeof(A.begin()) it=A.begin(); it != A.end(); ++it)
#define REP_S(i, str) for (char*i=str;*i;++i)
#define REP_L(i, hd, suc) for (int i=hd;i;i=suc[i])
#define REP_G(i, u) REP_L(i,hd[u],suc)
#define REP_SS(x, s) for (int x=s;x;x=(x-1)&s)
#define DO(n) for ( int ____n = n; ____n-->0; )
#define REP_2(i, j, n, m) REP(i, n) REP(j, m)
#define REP_2_1(i, j, n, m) REP_1(i, n) REP_1(j, m)
#define REP_3(i, j, k, n, m, l) REP(i, n) REP(j, m) REP(k, l)
#define REP_3_1(i, j, k, n, m, l) REP_1(i, n) REP_1(j, m) REP_1(k, l)
#define REP_4(i, j, k, ii, n, m, l, nn) REP(i, n) REP(j, m) REP(k, l) REP(ii, nn)
#define REP_4_1(i, j, k, ii, n, m, l, nn) REP_1(i, n) REP_1(j, m) REP_1(k, l) REP_1(ii, nn)

#define ALL(A) A.begin(), A.end()
#define LLA(A) A.rbegin(), A.rend()
#define CPY(A, B) memcpy(A, B, sizeof(A))
#define INS(A, P, B) A.insert(A.begin() + P, B)
#define ERS(A, P) A.erase(A.begin() + P)
#define LBD(A, x) (lower_bound(ALL(A), x) - A.begin())
#define UBD(A, x) (lower_bound(ALL(A), x) - A.begin())
#define CTN(T, x) (T.find(x) != T.end())
#define SZ(A) int((A).size())

typedef long long LL;
//typedef long double DB;
typedef double DB;
typedef unsigned uint;
typedef unsigned long long uLL;

/** Constant List .. **/ //{

const int MOD = int(1e9)+7;
const int INF = 0x3f3f3f3f;
const LL INFF = 0x3f3f3f3f3f3f3f3fLL;
const DB EPS = 1e-9;
const DB OO = 1e20;
const DB PI = acos(-1.0); //M_PI;

char g[1111][1111];

int main()
{
    #ifdef DoubleQ
    freopen("in.txt","r",stdin);
    #endif
    int n;
    while(~scanf("%d",&n))
    {
        for(int i = 1 ; i <= n ; i ++)
        {
            for(int j = 1 ; j <= n ; j ++)
            {
                g[i][j] = 'D';
            }
        }
        int k = n / 2 ;
        int tmp = k;
        int flag = 0;
        int next = 0;
        for(int i = 1 ; i <= n ; i ++ )
        {
            if(next && k == tmp + 1)
                break;
            if(k == 0)
            {
                flag = 1;
                next = 1;
                k++;
                continue;
            }
            if(flag == 1)
            {

                for(int j = 1 ; j <= k ; j ++)
                    g[i][j] = '*';
                for(int j = n ; j > n - k ; j --)
                    g[i][j] = '*';
                k ++;
            }
            else if(flag == 0)
            {
                for(int j = 1 ; j <= k ; j ++)
                    g[i][j] = '*';
                for(int j = n ; j > n - k ; j --)
                    g[i][j] = '*';
                k -- ;
            }
        }
        for(int i = 1 ; i <= n ; i ++)
        {
            for(int j = 1 ; j <= n ; j ++)
            {
                cout << g[i][j];
            }
            cout << endl;
        }
    }
}