Given some natural numbers n and k, my goal is to write a C program that outputs a number formed by every k-th digit of n. I wrote a program as follows:
#include <stdio.h>
#include <math.h>
#define MAX 100
void printDigit(int n, int k)
{
int arr[MAX];
int i = 0;
int j, r;
while (n != 0) {
r = n % (int)pow(10,k);
arr[i] = r;
i ;
n = n / pow(10,k);
}
for (j = i - 1; j > -1; j--) {
printf("%d ", arr[j]);
}
}
int main()
{
int n = 12345678;
int k = 2;
printDigit(n,k);
return 0;
}
My code outputs the same number but partitioned into substrings of length k. Why is that and how can I fix it so that I get the number I wanted?
CodePudding user response:
Your logic is too complicated, but if you want to stick to it:
int intpow(int x, int y)
{
int result = 1;
while(y--) result *= x;
return result;
}
void printDigit(int n, int k)
{
int arr[MAX];
int i = 0;
int j, r, powv;
while (n != 0) {
powv = intpow(10,k -1);
n /= powv;
if(!n) break;
r = n % 10;
arr[i] = r;
i ;
n = n / 10;
}
for (j = i - 1; j > -1; j--) {
printf("%d ", arr[j]);
}
}
CodePudding user response:
n % (int)pow(10,k)
is the low-order k
digits of n
. If you just want one digit, use n % 10
.
Since pow(10, k)
returns a double
, it might not be an exact integer. You should round it to the nearest integer to do proper division.
void printDigit(int n, int k)
{
int arr[MAX];
int i = 0;
int j, r;
int divisor = lround(pow(10, k));
while (n != 0) {
r = n % 10;
arr[i] = r;
i ;
n = n / divisor;
}
for (j = i - 1; j > -1; j--) {
printf("%d ", arr[j]);
}
}