0 or 1

I - 0 or 1

Time Limit:2000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Submit Status Practice HDU 2608

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

 3123 

 

Sample Output

 100 

Hint

Hint S(3) = T(1) + T(2) +T(3) = 1 + (1+2) + (1+3) = 8 S(3) % 2 = 0 

 

#include<cstdio>using namespace std;int main(){    int t;   scanf("%d",&t);    while(t--)    {        int n;        scanf("%d",&n);        int count=0;        for(int i=1;i<=n;i++)        {            if(i*i<=n)            count++;            else                break;        }        for(int j=1;j<=n;j++)        {            if(2*j*j<=n)                count++;                else                    break;        }        printf("%d\n",count%2);    }    return 0;}