How to check if the numbers smaller than an integer variable contain a specific digit

Question

I have a question: I’m supposed to build a program where when I enter an integer below a hundred, all numbers smaller than said integer and containing the digit “3″ appear on the screen (etc, if I enter 14, the numbers “3, 13″ should appear).

However, there’s something wrong with my code, please help! Thank you!

The code:

#include <stdio.h>

int main(int argc, const char * argv [])
{
    int wholenumber;
    printf("百以内の整数を入力してください\n");
    scanf_s("%d", &wholenumber);

    while(0 <wholenumber)
        {
            wholenumber--;

        while(0 < wholenumber){
        wholenumber = 3 %10;
        wholenumber = 3 /10;

         if (wholenumber == 3);
         {printf("%3d",wholenumber);}
    }
    }
    return 0;
}

"there’s something wrong with my code" - please be more specific — Karoly Horvath, Jul 01 '13 at 05:46
After taking the modulo and quotient of `wholenumber` (???), `wholenumber` will *never* be equal to 0. — Makoto, Jul 01 '13 at 05:48

score 1 · Answer 1 · edited Jul 01 '13 at 20:24

1

I have no idea what your intention was with this code:

 wholenumber = 3 % 10;
 wholenumber = 3 / 10;

First line sets the variable to 3, the second to 0.. which forces the program to exit from the loop.

edited Jul 01 '13 at 20:24

perreal

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answered Jul 01 '13 at 05:49

Karoly Horvath

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It does? I was hoping to be able to check if an the number three was in the integer wholenumber with that (I was looking at this http://stackoverflow.com/questions/4977456/how-to-check-if-a-int-var-contains-a-specific-number page) – user2537878 Jul 01 '13 at 06:23

score 1 · Answer 2 · answered Jul 01 '13 at 06:21

1

If x is an integer between 0 and 99, the following will check whether either of its digits is a 3:

if (x / 10 == 3 || x % 10 == 3) {
   ...
}

I leave the rest as an exercise for the reader.

answered Jul 01 '13 at 06:21

NPE

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mvv1277 · Answer 3 · 2013-07-01T09:47:52.270

Code:

#include<stdio.h>

int main()
{
    int i,num;
    int t1,t2;             // Variable to store data temporary
    printf("\nEnter the number : \n");
    scanf("%d",&num);

    for(i=0;i<num;i++)
    {
     t1= i/10;
     t2= i%10;

     if((t1==3) || (t2 ==3))   //check if number has 3 in it
       printf(" %d", i);

    }
return 0;
}

This is code which is required.Thanks to @amulous for pointing out mistake. Note: It is assumed that entered number is less than 100 as required by user who asked question.

perreal · Answer 4 · 2013-07-01T20:14:21.590

0

You can write this more efficiently, and this will work for any upper limit:

#include <stdio.h>

int main(int argc, const char * argv [])
{
    int wholenumber;
    printf("百以内の整数を入力してください\n");
    scanf("%d", &wholenumber);
    for (int tens = 1; tens < wholenumber; tens *= 10) {
        for (int i = 3 * tens; i < wholenumber; i+= 10*tens) {
            printf("%3d\n", i);
        }
    }
    return 0;
}

edited Jul 01 '13 at 20:14

answered Jul 01 '13 at 06:31

perreal

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3

it's more efficient for sure. but incorrect... it will never print `30`. – Karoly Horvath Jul 01 '13 at 06:34
I seriously hope that was a joke... :) – Karoly Horvath Jul 01 '13 at 06:36
I just can't believe you did that.. why is 30 special? – Karoly Horvath Jul 01 '13 at 10:06
@KarolyHorvath, I changed the code so that it is not handling 30 bending backwards :) – perreal Jul 01 '13 at 20:25

BLUEPIXY · Answer 5 · 2013-07-01T13:36:04.563

#include <stdio.h>

int contain3(int n){
    if(n == 0) return 0;
    return (n % 10 == 3) ? 1 : contain3(n/10);
}

int main(void){
    int i, wholenumber;
    printf("百以内の整数を入力してください\n");
    scanf_s("%d", &wholenumber);

    for(i=3;i<wholenumber;++i)
        if(contain3(i))
            printf("%3d", i);

    return 0;
}

#include <stdio.h>
#include <limits.h>

int main(void){
    int i, wholenumber;
    int data[]={3,13,23,30,31,32,33,34,35,36,37,38,39,43,53,63,73,83,93,INT_MAX};
    printf("百以内の整数を入力してください\n");
    scanf_s("%d", &wholenumber);

    for(i=0;data[i]<wholenumber;++i)
        printf("%3d", data[i]);

    return 0;
}

score 0 · Answer 6 · answered Jul 01 '13 at 18:40

OP solution is conceptionallly close but needs changes.

The while() loop test destroys wholenumber. The test for '3' containment should use a copy of wholenumber
the syntax of wholenumber = 3 /10; should be wholenumber_test /= 10;
other syntax errors.

void whole(int wholenumber, int digit) {
  while (0 < wholenumber) {
    int test = wholenumber;  // form test value
    while (0 < test) {
      if ((test%10) == digit) {
        printf("%3d\n", wholenumber);
        break;  // no need to test other digits
      }
      test /= 10;  // move onto next digit
    }
    wholenumber--;
  }
}

How to check if the numbers smaller than an integer variable contain a specific digit

6 Answers6