B. Present from Lena

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Vasya's birthday is approaching and Lena decided to sew a patterned handkerchief to him as a present. Lena chose digits from 0 to nas the pattern. The digits will form a rhombus. The largest digit n should be located in the centre. The digits should decrease as they approach the edges. For example, for n = 5 the handkerchief pattern should look like that:

          0
        0 1 0
      0 1 2 1 0
    0 1 2 3 2 1 0
  0 1 2 3 4 3 2 1 0
0 1 2 3 4 5 4 3 2 1 0
  0 1 2 3 4 3 2 1 0
    0 1 2 3 2 1 0
      0 1 2 1 0
        0 1 0
          0

Your task is to determine the way the handkerchief will look like by the given n.

Input

The first line contains the single integer n (2 ≤ n ≤ 9).

Output

Print a picture for the given n. You should strictly observe the number of spaces before the first digit on each line. Every two adjacent digits in the same line should be separated by exactly one space. There should be no spaces after the last digit at the end of each line.

Sample test(s)

input

2

output

    0
  0 1 0
0 1 2 1 0
  0 1 0
    0

input

3

output

      0
    0 1 0
  0 1 2 1 0
0 1 2 3 2 1 0
  0 1 2 1 0
    0 1 0
      0

解题说明：打印图形题，跳过，附别人代码

#include <iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;

int main()
{
    int n,m,i,j;
    scanf("%d",&n);
    m=2*n+1;
    for(i=0;i<m;i++)
    {
        if(i<n)
        {
            for(j=(n-i);j>0;j--)
            {
                printf("  ");
            }
            for(;j<i;j++)
            {
                printf("%d ",j);
            }
            for(;j>=0;j--)
            {
                if(j!=0)
                {
                    printf("%d ",j);
                }
                else
                {
                    printf("%d",j);
                }
            }
        }
        if(i==n)
        {
            for(j=0;j<n;j++)
            {
                printf("%d ",j);
            }
            for(;j>=0;j--)
            {
                if(j!=0)
                {
                    printf("%d ",j);
                }
                else
                {
                    printf("%d",j);
                }
            }
        }
        if(i>n)
        {
            for(j=(i-n);j>0;j--)
            {
                printf("  ");
            }
            for(j=0;j<(m-i-1);j++)
            {
                printf("%d ",j);
            }
            for(;j>=0;j--)
            {
                if(j!=0)
                {
                    printf("%d ",j);
                }
                else
                {
                    printf("%d",j);
                }
            }
        }
        printf("\n");
    }
    return 0;
}