Present from Lena

最新推荐文章于 2022-09-13 01:31:35 发布

原创最新推荐文章于 2022-09-13 01:31:35 发布

· 547 阅读

0 ·

版权

AC路漫漫专栏收录该内容

180 篇文章

订阅专栏

为庆祝Vasya的生日，Lena决定送给他一块带有图案的手帕作为礼物。手帕上的图案由数字组成，形成一个菱形，最大的数字位于中心，并且从中心向外逐渐减小。本篇介绍如何根据给定的最大数字n来确定手帕的最终图案。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Vasya's birthday is approaching and Lena decided to sew a patterned handkerchief to him as a present. Lena chose digits from 0 to n as 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

Output

Input

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

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

int main()  {
    int n,m,i,j;
    while(cin>>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;
}