Box of Bricks
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5994 Accepted Submission(s): 2599
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 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 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, 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.
Source
来源： <http://acm.hdu.edu.cn/showproblem.php?pid=1326>
这个就是个新手村史莱姆。相同的题有：HDOJ 1326、HDOJ 2088、POJ 1477、ZOJ 1251、UVA 594、UVA Live 5624。
题目大意是，有N堆砖块，每堆有h[i]个（也就是高度），问至少需要移动多少步砖块，使它变成一面平整的墙。
OK既然题目保证它是个墙，则砖的总数sum肯定能除尽N。求出平均高度ave后，累加高出平均高度的砖有多少块（低于平均高度的会被填平），就是需要移动的步数了。
1 #include <stdio.h>
2
3 int main()
4 {
5 int n, cse=1, h[51];
6 while(scanf("%d", &n), n){
7 printf("Set #%d\n", cse++);
8 int ave=0, res=0;
9 for(int i=0; i<n; i++) {
10 scanf("%d", h+i);
11 ave+=h[i];
12 }
13 ave/=n;
14 for(int i=0; i<n; i++)
15 if(h[i]>ave)
16 res+=h[i]-ave;
17 printf("The minimum number of moves is %d.\n\n", res);
18 }
19 return 0;
20 }