My program(LANG c)is in the following:
#include<stdio.h>
#include<stdlib.h>
#include<cmath>
int k[60000],value,i,n;
int main()
{
    scanf("%d",&n);
    for(i=1;i<=n;i++)
       scanf("%d",&k[i]);
    for(i=1;i<=n;i++)
    {
         if(k[i]==0) printf("0 ");
         else
         {
              value=(int)sqrt((k[i]-1)*2);
              if((value*(value+1))==2*(k[i]-1))
                 if(i==n)  printf("1\n");
                 else printf("1 ");
              else
                 if(i==n) printf("0\n");
                 else printf("0 ");
              }
    }
    return 0;
}

BUT it can't compile,said that
"
d9b0f699-b537-45eb-b286-ce7765096c87d9b0f699-b537-45eb-b286-ce7765096c87(16) : error C2668: 'sqrt' : ambiguous call to overloaded functionS:\checker\compile\vc10\include\math.h(589): could be 'long double sqrt(long double)' S:\checker\compile\vc10\include\math.h(541): or 'float sqrt(float)' S:\checker\compile\vc10\include\math.h(127): or 'double sqrt(double)' while trying to match the argument list '(int)'
"

BUT in my computer(Windows 7),I can compile and run it,and the answer is right.
So I don't know where is wrong?And I don't know how to modify it.Help me,please!

Edited by author 22.02.2013 15:09

length array int k[65535]
becouse
(1 ≤ N ≤ 65535)

your wrong is use (int) with sqrt.
template for SQRT is   sqrt(double).
you must to give for sqrt double variable, and if you need int variable, you must to appropriate value in int variable. Signs after a comma will be removed.
I am sorry for my bad English.

Edited by author 25.03.2013 20:14