codeforces 118B Present from Lena-优快云博客

本文链接：https://blog.youkuaiyun.com/zhang__liuchen/article/details/76177588

本文介绍了一个用于生成特殊手帕图案的C语言程序。该程序根据输入的整数n生成一个由数字0到n组成的菱形图案，中心为n，向边缘递减。文章提供了完整的代码实现，并通过示例展示了不同n值时的图案效果。

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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.

        Examples
      

          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

         
 
        

题意很简单，但是要准确考虑这些数字的每一个位置。

#include<stdio.h>
int main()
{
    int n,i,j;
    scanf("%d",&n);
    for(i=0; i<=n; i++)
    {
        for(j=0; j<(n-i)*2; j++)
            printf(" ");
        if(i==0)
            printf("0");
        else
        {
            int s=0;
            for(j=0; j<=i; j++)
                printf("%d ",s++);
            s--;
            for(j=0; j<i; j++)
            {
                if(j==i-1)
                    printf("%d",--s);
                else
                    printf("%d ",--s);
            }
        }
        printf("\n");
    }
    for(i=n-1; i>=0; i--)
    {
        for(j=(n-1)*2+1; j>=i*2; j--)
            printf(" ");
        if(i==0)
            printf("0");
        else
        {
            int s=0;
            for(j=0; j<=i; j++)
                printf("%d ",s++);
            s--;
            for(j=0; j<i; j++)
            {
                if(j==i-1)
                    printf("%d",--s);
                else
                    printf("%d ",--s);
            }
        }
        printf("\n");
    }
    return 0;
}