Codeforces 558C Amr and Chemistry（暴力+贪心）

最新推荐文章于 2019-03-25 10:13:44 发布

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最新推荐文章于 2019-03-25 10:13:44 发布

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探讨了在化学实验中，如何通过两种操作（体积翻倍或减半）使不同类型的化学物质达到相同的体积，以进行混合实验。分析了最优解的存在性和寻找方法，提供了一种算法实现思路。

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题面

Amr loves Chemistry, and specially doing experiments. He is preparing for a new interesting experiment.

Amr has n different types of chemicals. Each chemical i has an initial volume of ai liters. For this experiment, Amr has to mix all the chemicals together, but all the chemicals volumes must be equal first. So his task is to make all the chemicals volumes equal.

To do this, Amr can do two different kind of operations.

Choose some chemical i and double its current volume so the new volume will be 2ai
Choose some chemical i and divide its volume by two (integer division) so the new volume will be

Suppose that each chemical is contained in a vessel of infinite volume. Now Amr wonders what is the minimum number of operations required to make all the chemicals volumes equal?

Input

The first line contains one number n (1 ≤ n ≤ 105), the number of chemicals.

The second line contains n space separated integers ai (1 ≤ ai ≤ 105), representing the initial volume of the i-th chemical in liters.

Output

Output one integer the minimum number of operations required to make all the chemicals volumes equal.

Examples

Input

3
4 8 2

Output

Input

3
3 5 6

Output

Note

In the first sample test, the optimal solution is to divide the second chemical volume by two, and multiply the third chemical volume by two to make all the volumes equal 4.

In the second sample test, the optimal solution is to divide the first chemical volume by two, and divide the second and the third chemical volumes by two twice to make all the volumes equal 1.

题目链接

Codeforces_558C

题目简述

把n种化学物质变成相同体积，可做两种操作:①v*=2,②v/=2（整数除法）

假如v=31,那么可以哪些体积呢？

31、62、124、248……

15、30、60、120……

7、14、28、56……

3、6、12、24……

1、2、4、8……

所以要把每种化学物质可以到达的体积处理出来，处理过程中计算操作次数。(有些体积是v永远到达不了的，比如31到不了5)

还有一个结论：最终的体积不会超过n种化学物质初始体积的最大值maxi，因为（1-maxi）中会有更优解。

程序分析

node[i].legal：能到达体积i的化学物质种数

node[i].opera_num：操作次数

对于32这样的偶数来说，32->64,64->128,128->256……

那么32->16之后就没必要再乘2，因为之前到达过32。32->16,16->32,32->64……不是最优解，会浪费时间，适时break。

程序

#include<stdio.h>
#include<iostream>
using namespace std;
#define maxn 500005
struct Node
{
    int legal;
    int opera_num;
}node[maxn];
int a[maxn];
int maxi;

void init()
{
    for(int i=0;i<maxn;i++)
    {
        node[i].legal=0;
        node[i].opera_num=0;
    }
}

void deal(int x,int num)/**当前第num种化学物质**/
{
    for(int i=0;i<=100;i++)/**初始x到达当前x需要i步**/
    {
        int temp=x;
        for(int j=0;j<=100;j++)/**当前x->temp需要j步**/
        {
            if(temp>maxi||node[temp].legal==num)/**temp体积之前到达过**/
                break;
            if(node[temp].legal==num-1)/**前num-1种化学物质都能到达temp体积**/
            {
                node[temp].legal++;
                node[temp].opera_num+=i+j;
            }
            if(temp==0)
                break;
            temp*=2;
        }
        if(temp==0)
            break;
        x/=2;
    }
}

int main()
{
    init();
    int n;
    maxi=-1;
    scanf("%d",&n);
    for(int i=0;i<n;i++)
    {
        scanf("%d",&a[i]);
        maxi=max(a[i],maxi);
    }
    for(int i=0;i<n;i++)
    {
        deal(a[i],i+1);
    }
    int answer=100000000;
    for(int i=0;i<=maxi;i++)
        if(node[i].legal==n)/**n种物质都能到达**/
        {
            answer=min(answer,node[i].opera_num);
        }
    printf("%d\n",answer);
    return 0;
}