Print a different result than actual answer (Ciel and A-B Problem) - c

Question

Input:
An input contains 2 integers A and B.
Output:
Print a wrong answer of A-B. Your answer must be a positive integer containing the same number of digits as the correct answer, and exactly one digit must differ from the correct answer. Leading zeros are not allowed. If there are multiple answers satisfying the above conditions, anyone will do.
Code:
#include <stdio.h>
int no_of_zeroes(int x);
int main()
{
int a, b;
int res1, res2;
int n1, n2;
scanf("%d",&a);
scanf("%d",&b);
res1 = a - b;
res2 = res1 + 10;
n1 = no_of_zeroes(res1);
n2 = no_of_zeroes(res2);
if(res1 < 9) printf("%d",res1 + 1);
else if(res1 == 9) printf("%d",res1-1);
else if((n1 == n2) && (res1 > 9)) printf("%d",res2);
else if((n2 > n1) && (res1>9))
{
res2 = res2 - 20;
printf("%d",res2);
}
}
int no_of_zeroes(int x)
{
int count = 0;
while(x>0)
{
x = x / 10;
count++;
}
return count;
}
Error:
What different should i do, basically else if blocks are creating trouble.

Answer 1

Your answer must be a positive integer containing the same number of digits as the correct answer
So the answer should be a-b. But the statement also says
exactly one digit must differ from the correct answer
Since it didn't specify the position of the digit, changing the last digit only should give us the correct answer to this problem. And so counting the number of digits of the difference is redundant.
Now the issue is - how to change the last digit? It's simple. We add 1 to the difference.
But there is a catch! If the difference is 99 and we add 1 to it, the result will be 100. Here, not only we're changing more than one digit, but also the number of digits.
And so, all we have to do is subtract 1 from the difference if the last digit of the difference is 9.
And so, the if-else block should look something like this:
int diff = a - b;
if(diff%10 == 9) {
diff--;
}
else {
diff++;
}
Here's my full code:
#include <stdio.h>
int main(void) {
int a, b;
scanf("%d%d", &a, &b);
int diff = a - b;
if(diff%10 == 9) {
diff--;
}
else {
diff++;
}
printf("%d\n", diff);
return 0;
}
On a different note, the function no_of_zeroes(int x) will return 0 if x=0. But, it should return 1 under general circumstances. And so the function should be something like this:
int no_of_digits(int x)
{
/* Adding the following line should fix the issue */
if(x==0) return 1;
int count = 0;
while(x>0)
{
x = x / 10;
count++;
}
return count;
}

Answer 2

I think you are making this much more complicated than needed. All you need is to check the last digit of the correct result and then change it.
For 0 and positive numbers:
last digit is 0 : add 1
last digit is 1 : add 1
...
last digit is 8 : add 1
last digit is 9 : subtract 1
For negative numbers, you simply change the sign and handle it as the positive number. This can be done because -123 has the same digits as 123.
So the code can be:
void wrongCalc(int a, int b)
{
int res = a - b; // Calculate result
if (res < 0) res = -res; // Change sign if negative
int lastDigit = res % 10; // Find last digit
if (lastDigit == 9)
{
--res; // Subtract 1
}
else
{
++res; // Add 1
}
printf("%d - %d = %d (correct result is %d)\n", a, b, res, a-b);
}
Limitations:
1) The program doesn't handle the possible integer overflow in a-b
2) The program doesn't handle the possible integer overflow in res = -res;
3) The program doesn't handle the case where the correct result is INT_MAX