6 replies to this topic

[SpaceCowboy]

So as i mentioned i am rly bad at C, just started and i made a program, i think in a worst way it can be done but gives me the right answers, but can anyone write me a command line showing how many digits in sum are equal to 3
Here is my program dont laugh since this is the best one i've made so far. :w00t:

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

int main ()
{
int num, suma, a1, a2, a3, a4, a5, a6, max, I, n;
printf("Enter any number: \n");
scanf("%d", &num);
printf("Using a formula Sum=1+N+N*N (N=number of your choice)\n");
suma=1+num+num*num;
printf ("Sum: %d \n", suma);
a1=suma%10;
a2=suma/10%10;
a3=suma/100%10;
a4=suma/1000%10;
a5=suma/10000%10;
a6=suma/100000%10;
{
max=0;
for(I=0;I<n;I++);
{
if(max<a1)
{
max=a1;
}
if(max<a2)
{
max=a2;
}
if(max<a3)
{
max=a3;
}
if(max<a4)
{
max=a4;
}
if(max<a5)
{
max=a5;
}
if(max<a6)
{
max=a6;
}
}
printf("max=%d \n",max);
}
}

Edited by WingedPanther, 01 December 2010 - 06:39 PM.
add code tags (the # button)

[Alexander]

If you mean adapting your code to find all '3's in a sum given, you would need to modify your code to run in loops. For example
1) Get length of integer so you know how many times to extract each number
2) Run loop X times the length of the number
3) Loop to compare X results against '3'.

[SpaceCowboy]

Yeah thats what i want to get, I'll try, thou i have never even read about loop ;o

[mnirahd]

Hi,

You have to loop to determine the length of the number as 'n'; and then again using a loop upto 'n' times to get maximum digit!

[SpaceCowboy]

but how do i set the lengh of sum, i can set it if i choose any number but sum lengh is not up to me ;o

[Alexander]

You can use the same formula to grab the length of an integer through division, here is an example:

int intlen(int integer) {
    int i = 1;
    while ( (integer /= 10) > 0)
        i++;
    return i;
}

If you happened to go over the limit of the formula (such as integer /= 100000000) it will return a negative result, so we can safely break the loop and return the true count of numbers within an integer.

You can then of course store the individual lengths in an array instead of a1, a2, a3 .... a6 to make the program more flexible.

    //get length of sum
    length = intlen(suma);

    //prepare array to store results, a dynamic array of 'length' size
    int results[length]; 
    int divisor = 1;
    
    //store results in array
    for(i = 0; i <= length - 1; i++) {
        results[i] = suma / divisor % 10;
        divisor *= 10;
    }

notice how I set the divisor out of the loop, and increase it by 10 each time the loop happens so 10 -> 100 -> 1000 -> 10000 just like in your original equation, we then add it to the array dynamically using i as an index.

You can now iterate through the loop, here we will just print it, but you can check for 3's and create a count on each occurrence without much trouble:

    //now loop through array and print
    for(i = 0; i <= length - 1; i++) {
            printf("index %d = %d\n", i, results[i]);
    }

That code will work for any length number and result, as long as it is below integer max (~2.7 billion on 32 bit platforms, so you may wish to add checks for that if you wish)

A few more things you may wish to notice, such as array indexes starting at zero, so to iterate I use i = 0, and then length - 1 so I do not overrun the array and get a gibberish result from random memory, loops are quite easy once you get use to them!

[SpaceCowboy]

great info thenk you very much...