Sum of k-digits in ^n

[MoZak]

Hello guys how do i seperate a uknown number to find his digits?
I've done the work for a 4-digits number in ^4.
Now i should do the same for k-digits in ^n.
I should this k times???i tried with array but got confused...
:

#include <stdio.h>
#include <cstdlib>

// k =psifia n=dunami
main()
{
      int i,j,ar1,ar2,ar3,ar4,Din1,Din2,Din3,Din4,x,k,n,l,p_orio,d_orio,p;
      printf("Give  k and to n\n");
      scanf("%d %d",&k,&n);
      p_orio=1;
      d_orio=1;
      for (l=1; l<k; l++){
          p_orio=10*p_orio;
          }
          for(p=1; p<=k; p++){
                   d_orio=10*p_orio-1;
                         }
      printf("print %d %d\n",p_orio,d_orio);
      for(i=p_orio; i<=d_orio; i++){
      Din1=1;
      Din2=1;
      Din3=1;
      Din4=1;
               ar1=(i/1000);
                x=(i%1000);
                ar2=(x/100);
                x=(x%100);
                ar3=(x/10);
                x=(x%10);
                ar4=x;
                for (j=1; j<=n; j++){
                   Din1=Din1*ar1;
                   Din2=Din2*ar2;
                   Din3=Din3*ar3;
                   Din4=Din4*ar4;
                    }
                    if (i==Din1+Din2+Din3+Din4){
                    printf(number raised in %d  are: %d\n",n,i);
                    }
                    }
                    system("pause");
                    }

Help please xD

[CommonTater]

long long unknown;  // user input
int digits[16];
int idx = 0;

while (unknown > 0)
  { digits[idx] = unknown %10;
      unknown /= 10;
      idx++; }

At the end of the loop, each digit from unknown is in a separate array slot.

[MoZak]

The problem asks to print all number that are equal to their digits ^n
e.g 1634=1^4+6^4+3^4+4^4...So it should show all numbers from 1000 to 9999...
I did this and the programm stops working :

#include <stdio.h>
#include <cstdlib>

// k =psifia n=dunami
main()
{
      int i,j,ar1,ar2,ar3,ar4,Din1,Din2,Din3,Din4,x,k,n,l,p_orio,d_orio,p,dig[16],Din[16],sum,id=0;
      printf("Dwse to k kai to n\n");
      scanf("%d %d",&k,&n);
      p_orio=1;
      d_orio=1;
      for (l=1; l<k; l++){
          p_orio=10*p_orio;
          }
          for(p=1; p<=k; p++){
                   d_orio=10*p_orio-1;
                         }
      printf("emfanise %d %d\n",p_orio,d_orio);
      for(i=p_orio; i<=d_orio; i++){ //in case i add k=4,n=4 p_orio=1000 d_orio=9999
                    while (i>0){
                          dig[id]=i%10;
                          i/=10;
                          id++;
                          }
     
                for (j=1; j<=n; j++){
                Din[j]=1;
                }
                  for (j=1; j<=n; j++){
                
                  Din[j]=Din[j]*dig[id];
                  sum=sum+Din[j];
                    }
                    if (i==sum){
                    printf("oi arithmoi einai pou einai upswmenoi stin %di  dynami einai oi: %d\n",n,i);
                    }
                    }
                    system("pause");
                    }

[MoZak]

Any solution guys?

[laserlight]

Simplify the problem: write a program to print the digits of a number, with each digit separated by a space. You may print the digits in reverse order. What's your solution to this?

[MoZak]

I'll do and this buy why do i get an error??When i printf dig[id] it saws smth like that:
24422424424
200023
-0
4242422442
23230223
0
0
00
0
0
and then it stops working by getting an error

[laserlight]

What is your code for the simplified problem?

[MoZak]

I am stuck with this don't want to get more confused...please help me at this.

[laserlight]

Sorry, if you are so stuck that you cannot work on a simpler problem to get unstuck, then you must remain stuck, unless a flash of insight from your own mind saves the day. Law of the universe

You see, if you can write a loop to separate a number into its component digits, you can then apply the "raise to power of 4 and add the results together" part. What I am trying to get you to do is to leave out the other part so that you can see clearly how to do the "separate a number into its component digits" part. If you insist on doing both together, then what I can tell you is that... you should write a loop to separate a number into its component digits, raise each component digit to the fourth power and then add it to a running total. Easy

[MoZak]

:

 #include<stdio.h>
 main()
{
  int n  , i ;
  int dig[15] , top = 0;
  printf("Enter the number :");
  scanf("%d" , &n);
  while( n > 0 )
  {
        i = n % 10;
        dig[id] = i;
        n = n / 10;

  }
  while(dig)printf("%d " , dig[id]);
}

Is this right?????

[laserlight]

Goody. Unfortunately you have an infinite loop because dig never changes, and indeed it is an array so that is to be expected. Here's your program, adapted to work:

#include<stdio.h>

int main(void)
{
    int number;
    int digit;
    printf("Enter the number: ");
    scanf("%d", &number);
    while (number > 0)
    {
        digit = number % 10;
        printf("%d ", digit);
        number /= 10;
    }
    printf("\n");
    return 0;
}

Observe that I am not saving the digits in the array because I don't need to retrieve the digits later. Likewise, for the calculation that you are going to make, you don't need all the digits after the loop is done: you only need to result of the calculation.

(There is also an assumption that the number is positive, or at least non-negative, but this is probably reasonable.)

[MoZak]

Ok got it,but in my main programm what is wrong???and i get an error(I suck) -_-

[laserlight]

Wait. Here's one idea: extend the simplified version. Instead of printing the digit with a space, perform the calculation that you want to perform.

[CommonTater]

Ok got it,but in my main programm what is wrong???and i get an error(I suck) -_-

Laserlight is giving you extremely good advice here...

Are you even trying her examples?

[MoZak]

I am trying my friend be patient xDDD.
I understand how she is trying to help me but my mind can't reach hers xDD