Divide by three, multiply by two

本文介绍了一个算法问题,即如何通过重新排序一个包含特定数学操作(除以3或乘以2)的结果序列,来还原可能的操作流程。该问题要求输入序列能够按照特定规则重新组织,使得每个数都是前一个数的两倍或是三分之一。

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D. Divide by three, multiply by two
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output

Polycarp likes to play with numbers. He takes some integer number xx, writes it down on the board, and then performs with it n1n−1 operations of the two kinds:

  • divide the number xx by 33 (xx must be divisible by 33);
  • multiply the number xx by 22.

After each operation, Polycarp writes down the result on the board and replaces xx by the result. So there will be nn numbers on the board after all.

You are given a sequence of length nn — the numbers that Polycarp wrote down. This sequence is given in arbitrary order, i.e. the order of the sequence can mismatch the order of the numbers written on the board.

Your problem is to rearrange (reorder) elements of this sequence in such a way that it can match possible Polycarp's game in the order of the numbers written on the board. I.e. each next number will be exactly two times of the previous number or exactly one third of previous number.

It is guaranteed that the answer exists.

Input

The first line of the input contatins an integer number nn (2n1002≤n≤100) — the number of the elements in the sequence. The second line of the input contains nn integer numbers a1,a2,,ana1,a2,…,an (1ai310181≤ai≤3⋅1018) — rearranged (reordered) sequence that Polycarp can wrote down on the board.

Output

Print nn integer numbers — rearranged (reordered) input sequence that can be the sequence that Polycarp could write down on the board.

It is guaranteed that the answer exists.

这道题其实就是 按能膜3的数目由高到低排列 因为只有两种方式 一种是除以3一种是乘以2 所以能膜的3的数目肯定是依次递减的

并且 如果相邻的两者膜3的数目相等 那肯定是数小的在前面 数大的在后面 选择的是乘以2的方式

所以pair先比较除以3的数目 再比较乘以2的数目。

#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
typedef long long ll;
ll n,x;
pair<ll,ll>a[105];

int main(void)
{
    ll i,c,t;
    cin>>n;
    for(i=0;i<n;++i)
    {
        cin>>x;
        for(t=x,c=0;t%3==0;t/=3)c++;
        a[i]={-c,x};

    }
    sort(a,a+n);

    for(int i=0;i<n;++i)
    {
        cout<<a[i].first<<'*';
        cout<<a[i].second<<' ';
    }
    return 0;



}

1. Problem Description: A complex number is a number of the form a +bi, where a and b are real numbers and i is √-1 The numbers a and b are known as the real part and imaginary part of the complex number, respectively. You can perform addition, subtraction, multiplication, and division for complex numbers using the following formula: a+bi+c+di=(a+c)+(b+d)i a+bi-(c+di)=(a-c)+(b-d)i 第2页共2页 (a+bi)*(c+di)=(ac-bd)+(bc+ad)i (a+bi)/c+di)=(ac+bd)/c²+d²)+(bc-ad)i/(c²+d²) You can also obtain the absolute value for a complex number using the following formula: latbil=√a²+b (A complex number can be interpreted as a point on a plane by identifying the (a, b) values as the coordinates of the point. The absolute value of the complex number corresponds to the distance of the point to the origin, as shown in Figure 13.12b.) Design a class named Complex for representing complex numbers and the methods add, subtract, multiply, divide, abs for performing complex-number operations, and override toString method for returning a string representation for a complex number. The toString method returns a + bi as a string. If b is 0, it simply returns a. Provide three constructors Complex(a, b), Complex(a), and Complex(). Complex() creates a Complex object for number 0 and Complex(a) creates a Complex object with 0 for b. Also provide the getRealPart() and getlmaginaryPart() methods for returning the real and imaginary part of the complex number, respectively. Your Complex class should also implement the Cloneable interface. Write a test program that prompts the user to enter two complex numbers and display the result of their addition, subtraction, multiplication, and division. Here is a sample run: <Output> Enter the first complex number: 3.5 5.5 Enter the second complex number:-3.5 1 (3.5 + 5.5i) +(-3.5 + 1.0i)= 0.0 + 6.5 (3.5 + 5.5i)-(-3.5 + 1.0i)= 7.0 + 4.5i (3.5 + 5.5i)*(-3.5 + 1.0i) =-17.75 +-15.75i (3.5 + 5.5i) /(-3.5 + 1.0i)=-0.5094 +-1.7i |3.5 + 5.5il = 6.519202405202649 <End Output>
最新发布
06-09
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