二分贪心 U

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 50

Sample Output

Set #1The minimum number of moves is 5.

             这道题的贪心思路很简单，先求出平均高度，然后将每一数据高于均值的差值相加（高于跟低于的只算一类就行），所求和就是移动的最小砖数。

源代码如下：

#include<iostream>
using namespace std;
int main() 
{ int n,m,h,i,a[51],mi,t=1;
  while(cin>>n&&n!=0)
  { m=0;
   mi=0;
   for(i=0;i<n;++i)
     {
cin>>a[i];
      m+=a[i];
}
h=m/n;
  for(i=0;i<n;++i)
   if(a[i]>h)mi+=a[i]-h;
  cout<<"Set #"<<t++<<endl;
  cout<<"The minimum number of moves is "<<mi<<"."<<endl<<endl; 
  }
}

           需要注意的是每个输出数据后面还要求输出一个空行。