**ACM Volume V 591 - Box of Bricks**

**The Problem**

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 ≦ h

_{i}≦ 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**

6

5 2 4 1 7 5

0

**Sample Output**

Set #1

The minimum number of moves is 5.

**解題思考**

這一題要我們求出：把現有的這些方塊堆成每堆一樣高時，所需移動的最少方塊數。

其實，也只需要將每堆的平均方塊數(總積木數 / 堆)，再計算高於(或是低於也行)平均方塊數量的方塊數量總和，結果就出來了。

**參考解答(C++)**

#include <iostream> #include <iomanip> using namespace std; int main(void) { int c = 0; while (1) { c++; int n; cin >> n; if (n == 0) { break; } int average = 0; int *brick = new int[n]; for (int i = 0; i < n; i++) { cin >> brick[i]; average += brick[i]; } average /= n; int move = 0; for (int i = 0; i < n; i++) { if (brick[i] > average) { move += brick[i] - average; } } cout << "Set #" << c << endl; cout << "The minimum number of moves is " << move << "." << endl << endl; } #ifndef ONLINE_JUDGE system("pause"); #endif }

## 1 回覆:

這題我用了C#3.0的新語法：Lambda運算式

這是用來Functional Programming

http://kgame-blog.spaces.live.com/blog/cns!1A7962FEA74AB4CB!399.entry

程式中.Sum(m => m - average);

的m => m - average就是Lambda型式的匿名涵數

Functional Programming用於集合上的運算真是太強大了

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