# Can you make the code for this output?

Discussion in 'C' started by andyfeb, Sep 17, 2007.

1. ### andyfebNew Member

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Can you make the code for this output? (using looping, not gotoxy)

1.
Input: 4
Output:
1
234
234
1
Input: 5
Output:
1
234
56789
234
1

2.
Input: 5 (must odd)
1
2 2
3 1 3
4 2 2 4
5 3 1 3 5
4 2 2 4
3 1 3
2 2
1

2. ### andyfebNew Member

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I mean

1.
Input: 4
_1
234
_1
Input: 5
__1
_234
56789
_234
__1

2.
Input: 5
____1
___2_2
__3_1_3
_4_2_2_4
5_3_1_3_5
_4_2_2_4
__3_1_3
___2_2
____1

Note: _ is space, don't print it

3. ### DaWeiNew Member

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This is your assignment, right? YOU do it. Write some code and post for help with your problems. Before you do anything else, read the "Before you make a query" thread.

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I don't understand your first assignment or at least not the logic behind it.

5. ### andyfebNew Member

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Oh, I'm sorry. I newbie in this forum. I made the code, but I only can made it for 1 variable, I don't know how to make it in number counter.
Code:
```#include<stdio.h>
#include<conio.h>
void main()
{
int n,i,j,spasi;
printf("Input:");
scanf("%d",&n);
if(n%2==0)
{for(i=n/2;i>=1;i--)
{for(spasi=1;spasi<=i-1;spasi++)
printf(" ");
for(j=1;j<=2*(n/2-i)+1;j++)
printf("*");
printf("\n");}
for(i=1;i<=n/2;i++)
{for(spasi=1;spasi<=i-1;spasi++)
printf(" ");
for(j=1;j<=2*(n/2-i)+1;j++)
printf("*");
printf("\n");}}
else
{for(i=n/2+1;i>=1;i--)
{for(spasi=1;spasi<=i-1;spasi++)
printf(" ");
for(j=1;j<=2*(n/2+1-i)+1;j++)
printf("*");
printf("\n");}
for(i=1;i<=n/2;i++)
{for(spasi=1;spasi<=i;spasi++)
printf(" ");
for(j=1;j<=2*(n-i)-n;j++)
printf("*");
printf("\n");}}
getch();
}```

Last edited by a moderator: Sep 17, 2007

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