How do I write c program to generate 012345678910987654321, with these conditions

Question

1.00/5 (4 votes)

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1. Output should have 20 in nombers
2. program should contain only one loop or condition or ternary operator or anything
3. shouldn't initialize any number available in the series except 1
4. should not tested any nos available in the series (like i<10, i<=5)

[Update]
I wrote and tried many times as like below

C++

#include <stdio.h>
#include <stdlib.h>

int main()
{
    static int i,j;
    printf("%d",i);
    for(i=1,j=1;i<11,j<(1000/100);i++,j++)
    {
        printf("%d",i);

        printf("%d",10-j);
    }
    return 0;
}

Posted 19-Oct-15 22:46pm

Anupkumar_vj

Updated 15-Dec-17 9:12am

Patrice T

v3

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Comments

Krunal Rohit 20-Oct-15 4:50am

Homework!

-KR

Anupkumar_vj 21-Oct-15 13:30pm

Its was asked in job interview, finding solution but not getting. I want answer for next some other interview

Mohibur Rashid 20-Oct-15 18:02pm

Have you learned for loop yet?

Anupkumar_vj 21-Oct-15 13:30pm

Yea. I learnt for loop. But only one loop should be used

Philippe Mori 21-Oct-15 13:22pm

You cannot use comma operator to check 2 conditions. If i and j are always equal, there is no point to have 2 variables. You cannot have embedded loop using a single for... Well, you can simulate in using modulo and division operators...

Philippe Mori 21-Oct-15 13:32pm

Something like i - (i / 11) * 2 * (i-10) works.
Any number <= 10 give themself.
11 would give 11 - (11/11) * 2 * (11-10) = 9 (aasuming integer division).
12 would give 12- (12/11) * 2 * (12-10) = 8
Etc...

Anupkumar_vj 22-Oct-15 4:01am

You have given slight mathematical idea..
I used my own idea like this
#include <stdio.h>
#include <stdlib.h>

int main()
{
int i,j;
for(i=1,j=100/50;i<21;i++,j=j+2)
{
printf("%d",((i-1)-(i-1)/10*(j-22)));
}
return 0;
}
Will it satisfy rules ? yes, its giving output in codeblock..

Matt T Heffron 21-Oct-15 20:29pm

A common question in the various solutions below has to do with the "20 numbers".
Is that the count of digits in the resulting display (i.e., the 10 counts for 2)? (the solution you show has 21 digits)
Or is it the count of numbers used to make up the result (i.e., the 10 counts for 1)?

Anupkumar_vj 22-Oct-15 4:03am

You have given slight mathematical idea..
I used my own idea like this
#include <stdio.h>
#include <stdlib.h>

int main()
{
int i,j;
for(i=1,j=100/50;i<21;i++,j=j+2)
{
printf("%d",((i-1)-(i-1)/10*(j-22)));
}
return 0;
}
Will it satisfy rules ? yes, its giving output in codeblock comipler..

5 solutions

Solution 6

//I used my own idea and did like this

C++

#include <stdio.h>
#include <stdlib.h>

int main()
{
    int i,j;
    for(i=1,j=100/50;i<21;i++,j=j+2)
    {
     printf("%d",((i-1)-(i-1)/10*(j-22)));
    }
    return 0;
}

Will it satisfy rules ? yes, its giving output in codeblock comipler..

Posted 21-Oct-15 22:07pm

Anupkumar_vj

Updated 22-Oct-15 9:02am

v2

Comments

Patrice T 22-Oct-15 15:17pm

You can replace j=100/50 with j=2
the loop can be simplified this way

for(i=0,j=2;i<20;i++,j=j+2) {
    printf("%d",((i)-(i)/10*(j-22)));
    }

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OriginalGriff · Accepted Answer · 2015-10-19T22:59:00

We do not do your homework: it is set for a reason. It is there so that you think about what you have been told, and try to understand it. It is also there so that your tutor can identify areas where you are weak, and focus more attention on remedial action.

Try it yourself, you may find it is not as difficult as you think!

If you meet a specific problem, then please ask about that and we will do our best to help. But we aren't going to do it all for you!

So start by doing it manually, and work out how you do it.
Then you can look at translating that into code.

phil.o · Accepted Answer · 2015-10-19T23:01:00

1) Read your course carefully.
2) See how it is related to your exercise.
3) Write down on a paper how you are going to do things in peseudo-code.
4) Open your IDE, or a text editor if you prefer.
5) Begin to code according to point 3.
6) At some point, you may experience difficulty, you may then come back here ans ask a specific question on a specific issue.

Patrice T · Accepted Answer · 2015-10-20T10:54:00

Your code is confused !

Here is a first correction which gives you the first part of the wanted result.

C++

#include <stdio.h>
#include <stdlib.h>

int main() {
    static int i;
    for(i=0; i<11; i++) {
        printf("%d",i);
    }
    // Here is the place to make the second part of what you want
    return 0;
}

I let you complete the code for the end of the result.

Nota: it is a good habit to always initialise all your variables. you see that my code don't use uninitialised variables contrary to your code.

[Update]
Ok the piece of code do only half of the job. Here is a lead to solution:
Actual piece of code use f(x)= x
In text below:

x      -5 -4 -3 -2 -1  0  1  2  3  4  5
f(x)=x -5 -4 -3 -2 -1  0  1  2  3  4  5
?       5  4  3  2  1  0  1  2  3  4  5
?      -5 -4 -3 -2 -1  0 -1 -2 -3 -4 -5
?       5  6  7  8  9 10  9  8  7  6  5

line 1 is values of x
line 2 is actual f(x)
which mathematical function gives results of line 3 ?
What change gives results of line 4 ?
What change gives results of line 5 ?

milev yo · Accepted Answer · 2015-10-20T15:50:00

Solution 5

what about this?

C++

int recurPrint(int n){
    printf("%d",(9)-n);
    if(n){                   /*the only condition*/
        recurPrint(n-1);
    }
    printf("%d",(10)-n);
}
int main(){
    recurPrint(9);            /*the only initialization*/
    printf("\n");
}

Posted 20-Oct-15 15:50pm

milev yo

Updated 21-Oct-15 16:12pm

v3

Comments

Patrice T 21-Oct-15 2:16am

Just the loop is missing

Anupkumar_vj 21-Oct-15 13:37pm

I think printf returns 21 but output should have 20 in numbers.

Patrice T 21-Oct-15 14:54pm

See result as 0 1 2 3 4 5 6 7 8 9 10 9 8 7 6 5 4 3 2 1

Anupkumar_vj 22-Oct-15 4:05am

You have given slight mathematical idea..
I used my own idea like this
#include <stdio.h>
#include <stdlib.h>

int main()
{
int i,j;
for(i=1,j=100/50;i<21;i++,j=j+2)
{
printf("%d",((i-1)-(i-1)/10*(j-22)));
}
return 0;
}
Will it satisfy rules ? yes, its giving output in codeblock comipler..

How do I write c program to generate 012345678910987654321, with these conditions

5 solutions

Solution 1

Solution 2

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Solution 5

Solution 6

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