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 * 2P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
3 3
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>
看过我的代码之后，是不是觉得没问题呀，恩，反正当初我觉得就是对，但是，你可以把我的代码自己运行一下，测试数据随便取个奇数就ok啦，如3 23，哎，当初我测试的时候就没测试奇数，测试的全是偶数，XX还嘲笑我。后来得知超时主要是这几个原因：1.算法太简单2.效率太低，前面提到过，编程很讲究效率的3.程序遇到某种测试数据导致死循环。OK，下面看一下我提交后过了的代码吧。
#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;
}
这道题还是要从数的奇偶性出发。