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Box of Bricks
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 15307 Accepted Submission(s): 5087
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.
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.
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
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思路：
这输出方式也是学习了。
AC CODE：
#include<stdio.h>
#include<cstring>
#include<algorithm>
#define HardBoy main()
#define ForMyLove return 0;
using namespace std;
const int MYDD = 1103;int HardBoy {int n, h[MYDD], cnt = 0;while(scanf("%d", &n) && n) {if(cnt) puts("");//坑人的输出格式int ave = 0, ans = 0;for(int j = 0; j < n; j++) {scanf("%d", &h[j]);ave += h[j];}ave /= n;//平均值for(int j = 0; j < n; j++) {if(h[j] > ave) ans += (h[j]-ave);}printf("%d\n", ans);cnt++;}ForMyLove
}
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