HDU 3608 0 or 1（求一个数因子和的奇偶）

0 or 1

Time Limit: 6000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3460    Accepted Submission(s): 988

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

3123 

 

Sample Output

100
Hint
Hint S(3) = T(1) + T(2) +T(3) = 1 + (1+2) + (1+3) = 8     S(3) % 2 = 0 
 

 

Author

yifenfei

 

Source

奋斗的年代

 

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题目大意：

    输入一个数n，求1～n的数每个数因数和的和的奇偶性。

解题思路：

    参考了大神的博客

    因为因数一定能拆成质因数的乘积，而质因数除了2以外都是奇数，而2由2组成的因数的部分和一定为奇数，要想所有因数和为奇数。必须其它质因数的次数全为偶数，所以一定满足i*i==x || 2*i*i==x，所以我们只需要sqrt(n)的时间找到所有不大于n的和为1的数即可。

AC代码：

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;#define LL long longint main(){    int T;    scanf("%d",&T);    while(T--)    {        LL x;        bool ans=false;        scanf("%lld",&x);        for(LL i=1;i*i<=x;++i)        {            ans=!ans;            if(2*i*i<=x)                ans=!ans;        }        printf("%d\n",ans?1:0);    }        return 0;}