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最新推荐文章于 2021-09-20 22:11:39 发布

原创最新推荐文章于 2021-09-20 22:11:39 发布 · 1.9k 阅读

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#integer #output #numbers #input #each #math

ACM解题报告同时被 2 个专栏收录

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本文探讨了一个关于数学问题的解决方法，通过定义T(n)和S(n)来求解特定数学序列的问题，并提供了输入输出样例进行说明。算法通过打表方式优化求解过程，适用于快速获取特定范围内的序列和其模2后的结果。

0 or 1

Time Limit: 6000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 986 Accepted Submission(s): 204

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


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

题目不多解释了。

打表前50个元素如下

10011110111111100111111100000001111000000000000010.....

经过打表之后，发现转折点(0->1或者1->0)

有如下规律

1 1 平方根为:1
2 0
4 1 平方根为:2
8 0
9 1 平方根为:3
16 0 平方根为:4
18 1
25 0 平方根为:5
32 1
36 0 平方根为:6
49 1 平方根为:7
50 0
64 1 平方根为:8
72 0
81 1 平方根为:9
98 0
100 1 平方根为:10
121 0 平方根为:11
128 1
144 0 平方根为:12
162 1

其他非平方数的数均是由2与一个平方数相乘得到，如18=2*9,50=2*25，98=2*49

于是乎打表。把所有转折点打出来。

再现行查看给的n属于哪个范围即可！

CODE:

#include <iostream>
#include <algorithm>
using namespace std;
const __int64 inf=1<<31;
__int64 num[100000];
int main()
{
	__int64 i,j;
	num[0]=0;
	for(i=1;i<=46368;i++)
		num[i]=i*i;
	int cnt=i;
	for(i=0;i<=46368;i++)
		num[cnt++]=num[i]<<1;

	sort(num,num+cnt);
	int t,n;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d",&n);
		for(i=2;i<cnt;i++)
		{
			if(n>=num[i]&&n<num[i+1])
			{
				printf("%d\n",(i+1)&1);
				break;
			}
		}
	}
	return 0;
}