hdu1326

Box of Bricks

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 4373    Accepted Submission(s): 1959

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1326

Problem 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.
解题思路：
    题意是求为了让那些墙同样高，最少需要移动的砖块数。
    排个序，所有比平均数大的数的和就是最少需要移动的砖块数。
    最后····注意标点符号····输出时最后那个实心点····还有空行····不然很容易在这上边WA或者PE。
完整代码：
#include <functional>#include <algorithm>#include <iostream>#include <fstream>#include <sstream>#include <iomanip>#include <numeric>#include <cstring>#include <climits>#include <cassert>#include <complex>#include <cstdio>#include <string>#include <vector>#include <bitset>#include <queue>#include <stack>#include <cmath>#include <ctime>#include <list>#include <set>#include <map>using namespace std;typedef long long LL;typedef double DB;typedef unsigned uint;typedef unsigned long long uLL;/** Constant List .. **/ //{const int MOD = int(1e9)+7;const int INF = 0x3f3f3f3f;const LL INFF = 0x3f3f3f3f3f3f3f3fLL;const DB EPS = 1e-9;const DB OO = 1e20;const DB PI = acos(-1.0); //M_PI;int a[100001];int main(){    #ifdef ZH    freopen("in.txt","r",stdin);    #endif    int n;    int t = 1;    while(~scanf("%d",&n) && n)    {        int sum = 0;        for(int i = 0 ; i < n ; i ++)        {            scanf("%d",&a[i]);            sum += a[i];        }        int aver = sum / n;        sort(a , a + n);        int cnt = 0;        for(int i = n - 1 ; i >= 0 ; i --)        {            if(a[i] == aver)                break;            if(a[i] > aver)               cnt += a[i] - aver;        }        printf("Set #%d\n",t ++);        printf("The minimum number of moves is %d.\n\n",cnt);    }}