POJ 1477:Box of Bricks

POJ 1477:Box of Bricks

Box of Bricks

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

65 2 4 1 7 5 0

Sample Output

Set #1The minimum number of moves is 5.

Source

​​Southwestern European Regional Contest 1997​​

你  离  开  了  ，  我  的  世  界  里  只  剩  下  雨  。  。  。

#include int main(){ int n,i,s,aver,a[51],num,j=0; while(~scanf("%d", &n)) { if(n == 0) break; j++,s = 0,num = 0; for(i = 0; i < n; i++) { scanf("%d", &a[i]); s+=a[i]; } aver= s/n; for(i = 0; i < n; i++) if(a[i] > aver) num = num + a[i] - aver; printf("Set #%d\nThe minimum number of moves is %d.\n\n", j, num); } return 0;}

版权声明：本文内容由网络用户投稿，版权归原作者所有，本站不拥有其著作权，亦不承担相应法律责任。如果您发现本站中有涉嫌抄袭或描述失实的内容，请联系我们jiasou666@gmail.com 处理，核实后本网站将在24小时内删除侵权内容。