#include <stdio.h>
#include <math.h>
double stack[131073];
int main()
{
    int index = -1;
    while(scanf("%lf", &stack[++index]) != EOF);
    for(; index > 0;printf("%.4lf\n", sqrt(stack[--index])));
    return 0;
}

If the digit is negative?

Just use the same method, and append "i" to denote the imaginary number.

Hi, can this code be accepted?
The input value can be much larger than double value, even long type.

A strange fact:

[...]
double  b[100000000]; int i = 0;

int main(){
    while (scanf("%lf", &b[i++]) == 1); i--;
    while (i > 0)
        printf("%.4lf\n", sqrt((double)b[--i]));
    return 0;
}

Get AC
BUT:

[...]
long long  b[100000000]; int i = 0;

int main(){
    long long a;
    while (scanf("%lld", &b[i++]) == 1); i--;
    while (i > 0)
        printf("%.4lf\n", sqrt((double)b[--i]));
    return 0;
}

Gets WA 1.
I understand nothing in this life..=) I wonder where may be mistake?

Compiler G++ 4.7.2

lol 0<=A<=10^18 read description :P

Edited by author 21.11.2013 00:35

Edited by author 21.11.2013 00:35

Mingw gcc does not recognize %lld, use %I64u.
And now it does not work with %.4lf in printf, only with %.4f (in C++11 mode).

Well, it should be double not long long since the output is in real number