Problem(again)

**alperen1994** · 04-20-2009

i wrote this program,there are some problems

Code:

#include <stdio.h>


long int pow(int a,int b){
    int x=0,y=1;
    while(x<b){
    y*=a;
    x++;}
    return y;}


int rakam_top(long int a){
	int x=10,y=0;
	while((x<=a)&&(x%10==0)){
		if (x==10){y+=a%x;}
		y+=a/x;
		x*=10;}
	return y;}


int main(){
	int x,y;
	long int z;
	printf ("X raised to power of Y \n\n X=");
	scanf ("%i",&x);
	getchar();
	printf ("\n Y=");
	scanf ("%i",&y );
	getchar();
	z = pow(x,y);
	printf("\n %i raised to the power %i = %ld \n",x,y,z);
	printf("\n Digit sum= %i \n",rakam_top(z));
	getchar();
	return 0;}

when i declare x=2 and y=15
it finds the pow. right but Digit sum finds 3646/

when i declare x=2 and y=100
it return pow. 0

**Brafil** · 04-20-2009

Code:

int rakam_top(long int a)
{
    int result = 0;
    while (a / 10 > 0)
        result += a - (a / 10 * 10);

    return result;
}

2 ^ 100 = 10000000000000000........ in binary. The result is truncated, only the right bits are left. (Straight integer overflow, for every number 2 ^ n, where n is greater than a word (usually 32 bits))

**matsp** · 04-20-2009

1. Please alter your style to not put the ending brace on the last line of code - put it on it's own line on the next line [I personally prefer to have the starting brace on a separate line TOO, but at the very least the ending brace should be on it's own line].

2. What exactly does this part of your while statement do:

Code:

(x%10==0)

I'm not saying it's wrong, I just don't really understand it's meaning.

3. Are you sure this bit is right?

Code:

		y+=a/x;

--
Mats

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