[Zerojudge]c067: Box of Bricks

先找出平均數，就可以知道後來每一堆積木的個數。
將每一堆大於平均數的積木，將其個數減去平均數的數字(也就意味這堆要搬多少個盒子到別堆去)加起來就是答案。

[C++](8ms, 702KB)

	#include<iostream>
	using namespace std;
	int main()
	{
	int n;
	int boxh[55] = {0};
	int setnum = 0;
	while( cin >> n && n != 0 )
	{
	setnum++;
	int average = 0;
	int move = 0;
	for( int i = 0 ; i < n ; i++ )
	{
	cin >> boxh[i];
	average += boxh[i];
	}
	average /= n;
	for( int i = 0 ; i < n ; i++ )
	if( boxh[i] > average )
	move += boxh[i] - average;
	cout << "Set #" << setnum << endl;
	cout << "The minimum number of moves is " << move << ".\n\n";
	}
	return 0;
	}

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