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).
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
Hint
Hint S(3) = T(1) + T(2) +T(3) = 1 + (1+2) + (1+3) = 8
S(3) % 2 = 0
一开始做题的时候就理解错题意了，没有看到divided n，就理解为T(n)=1+n，(n>1),于是就用最简单的方法写的，但是结果超时了，后来才发现是理解错题意了，题意其实很简单，其中有规律的，我列了大量的数据才发现：
只有T[i*i]和T(i*i*2)为1,,其余的都为0，但是在写代码的过程中要注意是哪个在前哪个在后。所以：
#include <iostream>
#include <cstdio>
using namespace std;
int main()
{
int n;
int m,sum;
cin>>n;
while(n--)
{
scanf("%d",&m);
sum=0;
for(int i=1; i<=m; i++)
{
if(i*i*2<=m)
sum++;
if(i*i<=m)
sum++;
//if(i*i*2<=m)
//sum++;
else
break;
//cout<<sum<<endl;
}
//cout<<sum<<endl;
cout<<sum%2<<endl;
}
return 0;
}