hdu 2608 0 or 1(数论推理)

题目：http://acm.hdu.edu.cn/showproblem.php?pid=2608

0 or 1

Time Limit: 6000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2947    Accepted Submission(s): 795

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



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

 

Author

yifenfei

分析：这是我目前见过的最棒的数论题了。数论题就是这样，代码简洁而优雅。

/*
在所有的素因子中只有2是偶数，设number=2^p0*k1^p1*k2^p2*……*kn^pn，
那么T(n)=(1+2^1+2^2+……+2^p0)*(1+k1^1+……+k1^p1)*……*(1+kn^1+……+kn^pn)
因为(1+2^1+2^2+……+2^p0)%2=1，所以问题集中于(1+k1^1+……+k1^p1)*……*(1+kn^1+……+kn^pn) 
如果pi是一个奇数那么就有一个式子的结果是偶数（偶数个奇数是偶数）。
然后(1+k1^1+……+k1^p1)*……*(1+kn^1+……+kn^pn)%2=0.那么反过来想，所有的pi都是偶数的话
(1+k1^1+……+k1^p1)*……*(1+kn^1+……+kn^pn)%2=1（奇数个奇数是奇数，奇数乘以奇数还是奇数），
满足条件“所有的pi都是偶数”的数字(1+k1^1+……+k1^p1)*……*(1+kn^1+……+kn^pn)必然是另一个数字的平方。
所以number必然是里一个数字的平方（p0是偶数）或者一个数字平方的2倍（p0是奇数）
*/
#include <iostream>
#include <cstdio>
#include <cmath>
using namespace std;
int main()
{
    int t,n;
    cin>>t;
    while(t--){
        scanf("%d",&n);
        int res=(int)sqrt(n*1.0)+(int)sqrt(n*1.0/2);
        printf("%d\n",res%2);
    }
    return 0;
}