couldn't find out how to solve this algorithm problem - c

Question

I am beginner I tried so many times but I couldn't solve this problem I will be very pleased if you help me...
the question is:
Let x be an integer, and R(x) is a function that returns the reverse of the x in terms of its digits.
For example , if x:1234 then R(x)=4321.
Let’s call a positive integer mirror-friendly if it satisfies the following condition: 𝑥 + 𝑅(𝑥) = 𝑦^2 𝑤ℎ𝑒𝑟𝑒 𝑦 𝑖𝑠 𝑎𝑛 𝑖𝑛𝑡𝑒𝑔𝑒𝑟
Write a program that reads a positive integer as n from the user and prints out a line for each of the first n mirror-friendly integers as follows: x + R(x) = y^2
Example: If the user enters 5 as n, then the program should print out the following:
2 + 2 = 2^2
8 + 8 = 4^2
29 + 92 = 11^2
38 + 83 = 11^2
47 + 74 = 11^2
Here is the my code:
int
reverse(int num)
{
int reverse,
f,
i;
reverse = 0;
i = 0;
for (; i < num + i; i++) {
f = num % 10;
reverse = (reverse * 10) + f;
num /= 10;
}
return reverse;
}
int
sqrt(int n)
{
int i = 1;
int sqrt;
for (; i <= n; i++) {
sqrt = i * i;
}
return sqrt;
}
int
main()
{
int j = 1;
int main_num = 0;
for (; main_num <= 0;) {
printf("Please enter a positive integer: \n");
scanf_s("%d", &main_num);
}
int count = 0;
for (int i = 1; i <= main_num; i++) {
for (; j <= main_num; j++) {
if (j + reverse(j) == sqrt(?)) {
printf("%d + %d = %d\n", j, reverse(j), sqrt(?));
}
}
}
}

Answer 1

A few issues ...
sqrt does not compute the square root
reverse seems overly complicated
main_num (i.e. n from the problem statement) is the desired maximum count of matches and not the limit on x
Too many repeated calls to sqrt and reverse
No argument given to sqrt
The if in main to detect a match is incorrect.
sqrt conflicts with a standard function.
The variables you're using don't match the names used in the problem statement.
The printf didn't follow the expected output format.
Using a function scoped variable that is the same as the function is a bit confusing (to humans and the compiler).
Unfortunately, I've had to heavily refactor the code. I've changed all the variable names to match the names used in the problem statement for clarity:
#include <stdio.h>
#include <stdlib.h>
#ifdef DEBUG
#define dbgprt(_fmt...) printf(_fmt)
#else
#define dbgprt(_fmt...) do { } while (0)
#endif
int
reverse(int x)
{
int r = 0;
for (; x != 0; x /= 10) {
int f = x % 10;
r = (r * 10) + f;
}
return r;
}
int
isqrt(int x)
{
int y = 1;
while (1) {
int y2 = y * y;
if (y2 >= x)
break;
++y;
}
return y;
}
int
main(int argc,char **argv)
{
int n = -1;
--argc;
++argv;
if (argc > 0) {
n = atoi(*argv);
printf("Positive integer is %d\n",n);
}
while (n <= 0) {
printf("Please enter a positive integer:\n");
scanf("%d", &n);
}
int x = 1234;
dbgprt("x=%d r=%d\n",x,reverse(x));
int count = 0;
for (x = 1; count < n; ++x) {
dbgprt("\nx=%d count=%d\n",x,count);
// get reverse of number (i.e. R(x))
int r = reverse(x);
dbgprt("r=%d\n",r);
// get x + R(x)
int xr = x + r;
dbgprt("xr=%d\n",xr);
// get y
int y = isqrt(xr);
dbgprt("y=%d\n",y);
if (xr == (y * y)) {
printf("%d + %d = %d^2\n", x, r, y);
++count;
}
}
return 0;
}
Here is the program output:
Positive integer is 5
2 + 2 = 2^2
8 + 8 = 4^2
29 + 92 = 11^2
38 + 83 = 11^2
47 + 74 = 11^2
UPDATE:
The above isqrt uses a linear search. So, it's a bit slow.
Here is a version that uses a binary search:
// isqrt -- get sqrt (binary search)
int
isqrt(int x)
{
int ylo = 1;
int yhi = x;
int ymid = 0;
// binary search
while (ylo <= yhi) {
ymid = (ylo + yhi) / 2;
int y2 = ymid * ymid;
// exact match (i.e. x == y^2)
if (y2 == x)
break;
if (y2 > x)
yhi = ymid - 1;
else
ylo = ymid + 1;
}
return ymid;
}
UPDATE #2:
The above code doesn't scale too well for very large x values (i.e. large n values).
So, main should check for wraparound to a negative number for x.
And, a possibly safer equation for isqrt is:
ymid = ylo + ((yhi - ylo) / 2);
Here is an updated version:
#include <stdio.h>
#include <stdlib.h>
#ifdef DEBUG
#define dbgprt(_fmt...) printf(_fmt)
#else
#define dbgprt(_fmt...) do { } while (0)
#endif
// reverse -- reverse a number (e.g. 1234 --> 4321)
int
reverse(int x)
{
int r = 0;
for (; x != 0; x /= 10) {
int f = x % 10;
r = (r * 10) + f;
}
return r;
}
// isqrt -- get sqrt (linear search)
int
isqrt(int x)
{
int y = 1;
while (1) {
int y2 = y * y;
if (y2 >= x)
break;
++y;
}
return y;
}
// isqrt2 -- get sqrt (binary search)
int
isqrt2(int x)
{
int ylo = 1;
int yhi = x;
int ymid = 0;
// binary search
while (ylo <= yhi) {
#if 0
ymid = (ylo + yhi) / 2;
#else
ymid = ylo + ((yhi - ylo) / 2);
#endif
int y2 = ymid * ymid;
// exact match (i.e. x == y^2)
if (y2 == x)
break;
if (y2 > x)
yhi = ymid - 1;
else
ylo = ymid + 1;
}
return ymid;
}
int
main(int argc,char **argv)
{
int n = -1;
--argc;
++argv;
setlinebuf(stdout);
// take number from command line
if (argc > 0) {
n = atoi(*argv);
printf("Positive integer is %d\n",n);
}
// prompt user for expected/maximum count
while (n <= 0) {
printf("Please enter a positive integer:\n");
scanf("%d", &n);
}
int x = 1234;
dbgprt("x=%d r=%d\n",x,reverse(x));
int count = 0;
for (x = 1; (x > 0) && (count < n); ++x) {
dbgprt("\nx=%d count=%d\n",x,count);
// get reverse of number (i.e. R(x))
int r = reverse(x);
dbgprt("r=%d\n",r);
// get x + R(x)
int xr = x + r;
dbgprt("xr=%d\n",xr);
// get y
#ifdef ISQRTSLOW
int y = isqrt(xr);
#else
int y = isqrt2(xr);
#endif
dbgprt("y=%d\n",y);
if (xr == (y * y)) {
printf("%d + %d = %d^2\n", x, r, y);
++count;
}
}
return 0;
}
In the above code, I've used cpp conditionals to denote old vs. new code:
#if 0
// old code
#else
// new code
#endif
#if 1
// new code
#endif
Note: this can be cleaned up by running the file through unifdef -k

Answer 2

for(;i<num+i;i++)
is equal to
for(; 0<num;i++)
or for(; num;i++) if we are working with positive values only.
or even to while(num)
So, we don't need variable i in reverse function.
We don't need cycle at all in sqrt function. Just return n * n; is ok. But it is not sqrt then
The last cycle is too strange. At least variable j is not initialized.