HDU A Simple Task

最新推荐文章于 2024-08-28 08:54:03 发布

雪十月

最新推荐文章于 2024-08-28 08:54:03 发布

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分类专栏：基础题

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本文介绍了一种用于将正整数n分解为奇数o与2的幂次p相乘形式的算法，并提供了两种实现方式的C语言示例代码。通过该算法，可以有效地找出给定正整数n的奇数部分o及其指数p。

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Problem Description

Given a positive integer n and the odd integer o and the nonnegative integer p such that n = o2^p.

Example

For n = 24, o = 3 and p = 3.

Task

Write a program which for each data set:

reads a positive integer n,

computes the odd integer o and the nonnegative integer p such that n = o2^p,

writes the result.

Input

The first line of the input contains exactly one positive integer d equal to the number of data sets, 1 <= d <= 10. The data sets follow.

Each data set consists of exactly one line containing exactly one integer n, 1 <= n <= 10^6.

Output

The output should consists of exactly d lines, one line for each data set.

Line i, 1 <= i <= d, corresponds to the i-th input and should contain two integers o and p separated by a single space such that n = o2^p.

Sample Input

1
24

Sample Output

3 3

----------------------------------------------------------------
//遍历的方式，会超时，所以要用数学的方法
#include<stdio.h>
#include<math.h>
int main()
{
	//printf("%d\n",1<<2);
	int n,o,p,t,flag,temp;
	while(scanf("%d",&t)!=EOF)
	{
		while(t--)
		{
			scanf("%d",&n);
			flag=0;
			for(o=3;o<n;o+=2)
			{
				for(p=1;p<=sqrt(n)+1;p++)
					if(o*(int)pow(2,p)==n)
					{	flag=1; break;}
				
				if(flag)
				{	printf("%d %d\n",o,p); break;}
			}
		}
	}
	return 0;
}
PS : 用下面的方法；

/*


			for(p=0;p<=sqrt(n)+1;p++) p 的范围，可以通过观察的方式，找到大致范围，将p开方，可以大大减小，范围；
			{
				temp = (int)pow(2,p);// n/temp = o;
				if(n%temp==0&&(n/temp)&1)
					break;
			}
			printf("%d %d\n",n/temp,p);


*/