HDU 1339 A Simple Task

http://acm.hdu.edu.cn/showproblem.php?pid=1339

 

A Simple Task

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2207 Accepted Submission(s): 1231

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

 

Source

Central Europe 2001, Practice

 

Recommend

Ignatius.L

 

题目大意：求n = o * 2^p，给出n，求o和p。

分析：如果n是奇数，那么p的值必然是0，o的值为n。如果n是偶数，只需要不断地除2，一直到n为奇数，除2的次数为p的值。

代码如下：

#include<iostream>
using namespace std;
int main()
{
  // n=o*2^p    给n 求p o
  int T,n,o;
      scanf("%d",&T);
  while(T--)
  {
    scanf("%d",&n);
    if(n%2==1)
      printf("%d 0\n",n);
    else
    {
      o=0;
      while(n%2==0)
      {
        n=n/2;
        o++;
      }
      printf("%d %d\n",n,o);
    }
  }
  return 0;
}