Box of Bricks
Time limit:2000/1000 MS (java/others) Memory limit:65536/32768 K (java/others)
Total submission (s): 5457 Accepted Submission (s): 2358
Problem Descriptionlittle 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, should make all stacks the same height. Then you would has a real wall. ", she retorts. After a little con-sideration, Bob sees that she was right. So he sets out to rearrange the bricks, and one by one, such and all stacks is the same height afterwards. But since Bob was the lazy he wants to does this with the minimum number of bricks moved. Can help?
Inputthe input consists of several data sets. Each set begins with a line containing the number N of Stacks Bob have built. The next line contains n numbers, the Heights hi of the n stacks. Assume 1 <= n <= and 1 <= hi <= 100.
The total number of bricks is divisible by the number of stacks. Thus, it's always possible to rearrange the bricks such, and all stacks has the same height.
The input is terminated by a set starting with n = 0. This set should is processed.
Outputfor each set, first print the number of the and 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 has to be moved I n order to make all the stacks the same height.
Output a blank line after each set.
Sample Input
65 2 4 1 7 50
Sample Output
Set #1The Minimum number of moves is 5.
Sourcesouthwestern Europe 1997 Déjà vu! To find the average.
#include <iostream>using namespace Std;int main () {int n,aver;int ls[52];int s=0;while (cin>>n,n) {S++;int sum=0;for (int i=0;i<n;i++) {cin>>ls[i];sum+=ls[i];} aver=sum/n; int count=0;for (int j=0;j<n;j++) {if (ls[j]>aver) count+=ls[j]-aver;} cout<< "Set #" <<s<<endl;cout<< "the minimum number of moves is" <<count<< "." <<endl<<endl;;} return 0;}
Hangzhou Electric HDU ACM 1326 Box of Bricks
