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A Simple Task

          就像这道题的题目一样，真的是一个简单的题目，题意就是一句话：给一个数N求符合公式N = O * 2^P的 O 和 P。

下面我们就一起看一下题吧。

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

先给看一下我曾经提交很多次，但都超时的代码

<span style="font-size:24px;"><strong>#include<stdio.h>
int main()
{
	int d, n, o, p;
	while(scanf("%d",&d)==1)
	{
		for(int j=1;j<=d;j++)
		{
			scanf("%d",&n);
			o=1;
			p=1;
			if(n==1)
			{
				printf("0 1\n");
			}
			for(int i=2;j<n;i*=2)
			{
				
				if(n==o*i)
				{
					break;
				}
				else
				{
					o++;
					p++;
				}
			}
			printf("%d %d\n",o,p);
			
		}
	}
	return 0;
}</strong></span>

#include<stdio.h>
int main()
{
    int d,n,j;
    while(~scanf("%d",&d))
    {
    	for(int i=0;i<d;i++)
    	{
	    	scanf("%d",&n);
	    	j=0;
	    	while(n%2==0)
      		{   
       			j++;
          		n=n/2;     
          	}    
          	printf("%d %d\n",n,j);
	    }
    }
    return 0;
}

A Simple Task

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原文地址：http://blog.csdn.net/unusualnow/article/details/43796991