hdu 2608 0 or 1

最新推荐文章于 2020-02-12 17:52:53 发布

原创最新推荐文章于 2020-02-12 17:52:53 发布 · 702 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#math #G++

数论专栏收录该内容

1 篇文章

订阅专栏

本文介绍了一个有趣的数学问题，即如何计算S(n)=T(1)+T(2)+...+T(n)对2取模的结果，其中T(n)是所有能整除n的正整数之和。文章提供了一种高效的算法实现，并附上了C++代码。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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

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

通过打表会发现T(N)的一个特点，N=1,2,4,8,9,16,18,25,32,36,49,50,64....时T(N)%2==1，其余的都是0

结果见下表。
T(N)%2 1 1 0 1 0 0 0 1 1 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 1...

S(N) 1 0 0 1 1 1 1 0 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 0...
所以S(N)就是统计从 1-N 之间有若干好多个满足T(i)= 1的数

也可以证明参考http://blog.youkuaiyun.com/zuihoudebingwen/article/details/7976175

代码如下：

#include<cstdio>
#include<cmath>
#include<iostream>
using namespace std;
int main()
{
    int t,n,m;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        m=(int)sqrt(n)+(int)sqrt(n/2);
        printf("%d\n",m%2);
    }
    return 0;
}
注意要用G++交    c++不通过double转换成int