Tsurupeta - 1 month ago 11

C Question

I need your help on this problem , I've been trying to solve it all day but I can't reach a solution.

I just started studying C so I apologize if it is a stupid question, however

I am only allowed to use:

`if`

`for`

`do-while`

`while`

statements to solve the problem.

I need to check whether or not a given number can be written as the sum of two squares, I do not need to know which are these two squares, and I do not need to analyze cases when the number is 0 or 1. What I've managed to build till now is:

`unsigned int x;`

unsigned int q = 1;

printf("Enter a number : \n");

scanf("%u", &x);

unsigned int j = sqrt(x - (q*q));

if (x != 1 && x != 0)

for (q; (q*q) <= (x/2); q++)

if ((x - (q*q)) == (j*j))

printf("Given number is sum of two squares");

This one sometimes works and sometimes doesn't, for example it does work for

`65 (8^2+1^2)`

`90 (9^2+3^2)`

`(10^2+9^2)`

Do you have any idea how I can fix this?

Answer

Ok, so here's a pseudo code for a possible solution:

```
Input x;
int a;
int b;
for(a=0; a <= x; a++){
for(b=a; b <= x; b++){
if((a*a + b*b) == x){
Output is_solution;
}
}
}
```

In the nested loop, `b`

is assigned the value of `a`

to avoid checking the same sum of squares more than once.

Converting this to C, should look like this:

```
unsigned int x;
unsigned int a;
unsigned int b;
unsigned int is_solution = 0;
printf("Enter a number : \n");
scanf("%u", &x);
for(a=0; a<=x; a++){
for(b=a; b<=x; b++){
if((a*a + b*b) == x){
printf("Given number is sum of two squares");
}
}
}
```

I'm a bit rusted in C, hopefully it doesn't have any nasty errors.

Source (Stackoverflow)

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