HDU 2608 0 or 1 规律

点击打开链接

0 or 1

Time Limit: 6000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2258    Accepted Submission(s): 555

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

都能在网上找到规律就是：凡是“能够被完全开方”或者“被2整除后能够完全被开方”的数，它的(T(N) % 2)都是1。

从此我的博客里面多了一个规律的分组。

//15MS228K#include<stdio.h>#include<math.h>int main(){    int t;    scanf("%d",&t);    while(t--)    {        int n,count=0;        scanf("%d",&n);        for(int i=1;i<=n;i++)        {            if(i*i*2<=n)count++;            if(i*i<=n)count++;            else break;        }        printf("%d\n",count%2);    }    return 0;}