0 or 1

Problem Description
Solving problem is a interesting thing. Yifenfei like to slove different problem,because he think it is a way let him more intelligent. But as we know,yifenfei is weak in math. When he come up against a difficult math problem, he always try to get a hand. Now the problem is coming! Let we
define T(n) as the sum of all numbers which are positive integers can divied n. and S(n) = T(1) + T(2) + T(3)…..+T(n).

Input
The first line of the input contains an integer T which means the number of test cases. Then T lines follow, each line consists of only one positive integers n. You may assume the integer will not exceed 2^31.

Output
For each test case, you should output one lines of one integer S(n) %2. So you may see the answer is always 0 or 1 .

Sample Input
3
1
2
3

Sample Output
1
0
0

 #include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<iostream>
#define ll long long
using namespace std;
int main()
{
    int t,n,m;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        m=(int)sqrt(n)+(int)sqrt(n/2);
        printf("%d\n",m%2);
    }
    return 0;
}

#include
#include
#include
#include
#include
#define ll long long
using namespace std;
int main()
{
int t,n,m;
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
m=(int)sqrt(n)+(int)sqrt(n/2);
printf("%d\n",m%2);
}
return 0;
}