1177: 零起点学算法84——Box of Bricks

Time Limit: 1 Sec Memory Limit: 32 MB 64bit IO Format: %lld
Submitted: 769 Accepted: 268
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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

Source

零起点学算法

 1 #include<stdio.h>
 2 int main(){
 3     int n,a[50];
 4     while(scanf("%d",&n)!=EOF&&n!=0){
 5         int s=0;
 6         for(int i=0;i<n;i++){
 7             scanf("%d",&a[i]);
 8             s+=a[i];
 9         }
10         int ave,k=0;
11         ave=s/n;
12         for(int i=0;i<n;i++){
13             if(a[i]>ave) k+=a[i]-ave;
14         }
15         printf("%d\n\n",k);
16     }
17     return 0;
18 }

Output a blank line between each set. 输出每组之间的空行。不懂英文真坑爹。

时间： 2024-07-31 02:42:30

1177: 零起点学算法84——Box of Bricks