B - Box of Bricks

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本文详细解析了B-BoxofBricks问题，这是一个关于通过最少移动次数使所有砖块堆高度相等的算法挑战。文章提供了两种编码解决方案，一种采用暴力循环法，另一种则利用平均值策略优化效率。

B - Box of Bricks

Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u

Description

Little Bob likes playing with his box of bricks. He puts the bricks one upon another and builds stacks of different height. “Look, I've built a wall!”, he tells his older sister Alice. “Nah, you should make all stacks the same height. Then you would have a real wall.”, she retorts. After a little consideration, Bob sees that she is right. So he sets out to rearrange the bricks, one by one, such that all stacks are the same height afterwards. But since Bob is lazy he wants to do this with the minimum number of bricks moved. Can you help?

Input

The input consists of several data sets. Each set begins with a line containing the number n of stacks Bob has built. The next line contains n numbers, the heights hi of the n stacks. You may assume 1≤n≤50 and 1≤hi≤100.

The total number of bricks will be divisible by the number of stacks. Thus, it is always possible to rearrange the bricks such that all stacks have the same height.

The input is terminated by a set starting with n = 0. This set should not be processed.

Output

For each set, print the minimum number of bricks that have to be moved in order to make all the stacks the same height.
Output a blank line between each set.

Sample Input

6 5 2 4 1 7 5 0

Sample Output

题意分析:

用最少的移动次数使所有的砖块等高。

解题思路：

我的思路就是最暴力的方法，每次循环求出最高的和最矮的，然后将最高的砖块移动到最矮的上，循环到最后最高的高度和最矮的高度相差小于等于1的时候循环结束，输出移动次数。本次最大的坑点在于最后的输出格式，对于输出格式有严格的要求，最后一个输出样例不能被换行。

编码：

#include <bits/stdc++.h>

using namespace std;

int main()
{
    int n;
    int flag = 1;
    while(cin >> n)
    {
        if (n == 0)
        {
            break;
        }
        else
        {
            int num[101];
            int ans = 0;
            for (int i = 0; i < n; i++)
            {
                cin >> num[i];
            }
            int maxnum = -999;
            int minnum = 999999;
            int maxloc = -1;
            int minloc = -1;
            while (1)
            {
                maxnum = -999;
                minnum = 999999;
                maxloc = -1;
                minloc = -1;
                for (int i = 0; i < n; i++)
                {
                    if (num[i] > maxnum)
                    {
                        maxnum = num[i];
                        maxloc = i;
                    }
                    if (num[i] < minnum)
                    {
                        minnum = num[i];
                        minloc = i;
                    }
                }
                if (maxnum - minnum <= 1)
                {
                    break;
                }
                else
                {
                    ans++;
                    num[minloc]++;
                    num[maxloc]--;
                }
            }
            if (flag == 0)
            {
                cout << endl;
                cout << ans << endl;
            }
            else 
            {
                cout << ans << endl;
                flag = 0;
            }
        }
    }
    return 0;
}

编码2：

#include <iostream>
using namespace std;
int main()
{
    int n;
    int flag = 1;
    while (cin >> n)
    {
        if (n <= 0)
        {
            break;
        }
        else
        {
            int a[1001];
            int sum = 0;
            for (int i = 0; i < n; i++)
            {
                cin >> a[i];
                sum += a[i];
            }
            int temp;
            if (sum % n == 0)
            {
                temp = sum / n;
            }
            else
            {
                double avg = (double)sum/(double)n;
                temp = (int)avg;
                temp++;
            }
            int ans = 0;
            for (int i = 0; i < n; i++)
            {
                if (a[i] > temp)
                {
                    ans += (a[i] - temp);
                }
            }
            if (flag == 0)
            {
                cout << endl;
                cout << ans << endl;
            }
            else 
            {
                cout << ans << endl;
                flag = 0;
            }
        }
    }
    return 0;
}

最后：

这道题应该有更为简单的做法，在之前的训练中我也解决过这道题，当时用的是求平均值的办法，然后对于大于平均值的砖块移动，这样的效率会更高。