HDOJ 2088

本文主要是介绍HDOJ 2088，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！

Box of Bricks

Problem 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
5

Author
qianneng

Source
冬练三九之二

解题思路
大致的可以想象成，将高过平均值的砖块放置到低于平均值的那条上。根据题意，这些砖块总是会有一个等于整数的均值，这样就方便多了，只需要计算有哪些砖块高于均值即可。

AC

#include<stdio.h>
using namespace std;
int main() {int hi[55],n;while (scanf("%d", &n) != EOF && n != 0) {int cnt = 0,avg;for (int i = 0; i < n; i++) {scanf("%d", &hi[i]);cnt += hi[i];}avg = cnt / n;cnt = 0;for (int i = 0; i < n; i++) {if (hi[i] > avg)cnt += hi[i] - avg;}printf("%d\n", cnt);}return 0;
}

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