I am trying to solve this tutorial practice question that doesn't have an answer that I can check my code against. The goal is to write a program to display numbers whose digits are 2
greater than the corresponding digits of the entered number. So if the number input is 5656
then the output number should be 7878
. I have figured out how to separate each number and add them, but I can't seem to get them to print in a four-digit sequence.
#include <stdio.h>
int main ()
{
int n, one, two, three, four, final;
scanf("%d", &n);
one = (n / 1000);
n = (n % 1000) + 2;
two = (n / 100) + 2;
n = (n % 100) + 2;
three = (n / 10) + 2;
n = (n % 10) + 2;
four = (n / 1) + 2;
n = (n % 1) + 2;
final = (one * 1000) + (two * 100) + (three * 10) + four;
printf("%d", final);
return 0;
}
#include <stdio.h>
int main()
{
int n,a[4], final;
scanf("%d", &n);
for(int i=3;i>=0;i--)
{
a[i]=n%10+2;
n/=10;
}
final = (a[0] * 1000) + (a[1] * 100) + (a[2] * 10) + a[3];
printf("%d", final);
return 0;
}
Below function works with N number of digits. Idea is to extract each digit from the input number and add its decimal position.
#include <stdio.h>
int power(int x, int y)
{
int res = 1;
for (;y>0;y--)
{
res *=x;
}
return res;
}
int main ()
{
int n;
scanf("%d", &n);
int sum = 0;
int i=0;
while(n>0)
{
sum += ((n%10) +2)*power(10,i);
i++;
n /=10;
}
printf("%d", sum);
return 0;
}
Another idea:
char str[10]; // enough to contain an int as string + 1
char *s = str+sizeof(str); // points to last char + 1
int n;
scanf("%d", &n);
*--s = 0; // terminate the string
while(n) {
*--s = (((n % 10)+2)%10) + '0'; // write a char from the end
n /= 10;
}
printf("%s\n", s);
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