# B. Present From Lena

Source: Internet
Author: User
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 0NAs
The pattern. The digits will form a fig. The largest DigitNShocould be located in the center. The digits shoshould decrease as they approach
The edges. For example,NRows = rows 5 the handkerchief pattern shoshould 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 00 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 givenN.

Input

The first line contains the single integerN(2 cores ≤ CoresNLimit ≤ limit 9 ).

Output

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

Sample test (s) Input
`2`
Output
`    0  0 1 00 1 2 1 0  0 1 0    0`
Input
`3`
Output
`      0    0 1 0  0 1 2 1 00 1 2 3 2 1 0  0 1 2 1 0    0 1 0      0`

Solution Description: print the graphic questions, skip, and attach other people's code.

`#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;}`
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