Write a nested loop to create following pattern

[tienzyee]

I have to write a nested loop using for statement to produce the following pattern if the user input is 5

Please enter an integer > 5 (this will be using printf and scanf to get the integer)

1
121
12321
1234321
123454321


and also another program (different coding) to produce the following pattern if user input is 5:


1
2 2
3 3
4 4
55555555

[MitchMitchell]

I have to write a nested loop using for statement to produce the following pattern if the user input is 5

Please enter an integer > 5 (this will be using printf and scanf to get the integer)

1
121
12321
1234321
123454321


and also another program (different coding) to produce the following pattern if user input is 5:


1
2 2
3 3
4 4
55555555

Do you need help with printing or the loops?

The first loop is a simple for loop (i=1;i<=input;i ++)
{j=input-1;
print i;
if i=input
{ while(i>0)
if(j>0)
{print j;
--j;}
insert end of line;
}

You get the idea. Next one is just a simple variation.

Mitch

[tawanda]

Hie, for the first program you can use this code:
int i,j;
for(i=1;i<=5;i+ +)
{
{
for(j=1;j<=i;j+ +)
printf("%d",j);
for(j=(i-1);j>0;j--)
printf("%d",j);
}
printf("\n");
}

[tienzyee]

I'm so sorry, actually i mean the looping, not print the numbers, and also the message earlier doesnt allow me to type space in between....i want the following output using while loop statement....pl ease help...thank you

######1######
#####2#2#####
####3###3####
###4#####4###
##555555555##

in which # symbol is actually space....and 5 is the input


and also this output (for another different program), input also 5

######1######
#####121#####
####12321####
###1234321###
##123454321##

[D_C]

I'm not sure what is supposed to happen for numbers greater than nine. This does the simpler case (with spaces in between).

int i = 1;
int count = 0;
while(i <= input)
{
  while(count < ((2*input)+3))
  {
    if( ... )
      print a space
    else if( ... )
           print count
         else if ( ... )
           print either input or count
              else
                print a space

    count++;
  }
  i++;
}

[tienzyee]

i'm so sorry but i'm just starting to learn c in which i dun really understand the code that you gave, can someone make it simpler?

[micangello]

/* ###1### in this case input is 4 when i=0 u have 3 spaces
** ##2#2## when i=1 u have 2 spaces
** #3###3# when i=2 u have 1 space
** 4#####4 when i=3 u have 0 space
*/








#include<stdio. h>
main()
{
int i, j, numspace, input;
printf(" Type a number ");
scanf("%d", &input);
i = 0;


while(i < input)
{
numspace = input - ( i + 1); // in the first line when input is 4 and i is 0
// u have 4 - (1 + 0 ) spaces
// now a loop for printing spaces
for( j = 0; j< numspace; j++)
printf(" "); // this prints spaces
// now to print the number... after printing the spaces u simply print the number
// whis is ( i + 1 ) since on the first line.. the number is 1 and i is 0
// on the 2nd line where i is one.. the number is 2 .. that is ( i + 1 ) and so on
// so we have the following
printf("%d", (i + 1) );
// now.. to print the other number on the line...
// note that on the second line u have this sequence 2#2 one space btween them where i = 1
// and on the 3rd line........... ............... ... 3###3 3spaces between them where i = 2
// soo.. we have 2*i -1 spaces beween the number
// for i = 1 ... 2*1 - 1 = 1 for i = 2.... 2*2 - 1 = 3
// so.. we loop again for the other spaces
// but first we define the number of spaces
if(i>0) // since on the first line u only have one number
{
numspace = 2*i -1;
for(j=0; j<numspace; j++)
printf(" ");
printf("%d", i+1);
}


// almost done.. now we have to print a newline and increment i
printf("\n");
i++;

}
}

[tienzyee]

yeah that works but what about 5 ?the last line must print all 555555....

what about coding this program with only for statements? is it possible?

[tienzyee]

erm...can someone answer me please...