hdu 2608

0 or 1

Time Limit: 6000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

[align=left]Problem Description[/align]
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).

 

[align=left]Input[/align]
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.
 

[align=left]Output[/align]
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 .
 

[align=left]Sample Input[/align]

3
1
2
3

 

[align=left]Sample Output[/align]

1
0
0

HintHint S(3) = T(1) + T(2) +T(3) = 1 + (1+2) + (1+3) = 8
S(3) % 2 = 0
解题思路：这道题目我压根没思路，看了别人的解题报告。AC:

/*
分析:假设数n=2^k*p1^s1*p2^s2*p3^s3*...*pi^si;//k,s1...si>=0,p1..pi为n的素因子
所以T
=(2^0+2^1+...+2^k)*(p1^0+p1^1+...+p1^s1)*...*(pi^0+pi^1+...+pi^si);
显然(2^0+2^1+...+2^k)%2=1,所以T
是0或1就取决于(p1^0+p1^1+...+p1^s1)*...*(pi^0+pi^1+...+pi^si)
而p1...pi都是奇数(除2之外的素数一定是奇数),所以(pi^0+pi^1+...+pi^si)只要有一个si为奇数(i=1...i)
则(pi^0+pi^1+...+pi^si)%2=0,则T
%2=0//若si为奇数,则pi^si+1为偶数,pi^1+pi^2+...+pi^(si-1)为偶数(偶数个奇数和为偶数)
所以要T
%2=1,则所有的si为偶数，则n=2^(k%2)*m^2;//m=2^(k/2)*p1^(s1/2)*p2^(s2/2)*...*pi^(si/2)
所以只要n为某个数的平方或者某个数的平方和则T
%2=1,只要统计n的个数即可
*/
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<string>
#include<queue>
#include<algorithm>
#include<map>
#include<cmath>
#include<iomanip>
#define INF 99999999
using namespace std;

const int MAX=10;

int main(){
int t,n;
cin>>t;
while(t--){
cin>>n;
int sum=(int)sqrt(n*1.0)+(int)sqrt(n*1.0/2);
cout<<sum%2<<endl;
}
return 0;
}