A Simple Task

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本文解析了一道名为ASimpleTask的ACM编程题，题目要求输入一个正整数n，然后将其分解为一个奇数o乘以2的p次方的形式，并输出o和p的值。文中提供了一个C++实现的示例程序。

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): 4370 Accepted Submission(s): 2400

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

依据上述一个数只有唯一分解形式

#include <cstdio>
#define Max 1000000
int o[Max+10],p[Max+10];
void get_ans(void){
    for(int i=1;i<=Max;i++)
    {
        int x=2;
        int sum=0;
        while(i%x==0){x*=2;sum++;}
        x/=2;
        p[i]=sum;
        o[i]=i/x;

    }
}
int main()
{
    int t;
    get_ans();
    scanf("%d",&t);
    while(t--){
        int n;
        scanf("%d",&n);
        printf("%d %d\n",o[n],p[n]);
    }
    return 0;
}