uva 591 Box of Bricks（求平均值）_sample input 6 5 2 4 1 7 5 0 sample output set #1 -优快云博客

本文链接：https://blog.youkuaiyun.com/keshuai19940722/article/details/9664253

这篇博客讲述了小鲍勃如何将不同高度的砖堆重新排列，使其高度相同，以此建造一堵真正的墙。他希望以最小的砖块移动数量来完成，读者将学习如何计算最小移动数量的方法，即先计算平均高度，然后用高于平均值的堆的高度减去平均值，所有差值之和即为答案。输入和输出格式以及样例数据也包含在内。

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Box of Bricks

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 con- sideration, 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 h _i of the n stacks. You may assume $1 Ÿ\le n \leŸ 50$ and $1 \leŸ h_i Ÿ\le 100$ .

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, first print the number of the set, as shown in the sample output. Then print the line `` The minimum number of moves is k .'', where k is the minimum number of bricks that have to be moved in order to make all the stacks the same height.

Output a blank line after each set.

Sample Input

6
5 2 4 1 7 5
0

Sample Output

Set #1
The minimum number of moves is 5.

题目大意：把n堆砖移成相同高度需要移动多少块砖。

解题思路：先求平均值，再用比平均值大的数减去平均值，所有差的和就是所求值。

#include<stdio.h>
#include<string.h>

#define N 55

int main(){
	int n, t = 1;
	int num[N];
	while (scanf("%d", &n), n){

		// Init.
		memset(num, 0, sizeof(num));
		int sum = 0, aver;

		// Read.
		for (int i = 0; i < n; i++){
			scanf("%d", &num[i]);
			sum += num[i];
		}
		aver = sum / n;
		sum = 0;

		for (int i = 0; i < n; i++){
			if (num[i] < aver)
				sum = sum + aver - num[i];
		}

		printf("Set #%d\n", t++);
		printf("The minimum number of moves is %d.\n\n", sum);
	}
	return 0;}