Posted on 2010-08-07 18:08
MiYu 阅读(827)
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ACM ( 水题 ) 、
ACM ( 杂题 )
MiYu原创, 转帖请注明 : 转载自 ______________白白の屋
题目地址:
http://acm.hdu.edu.cn/showproblem.php?pid=2088
题目描述:
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 consideration, 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, print the minimum number of bricks that have to be moved in order to make all the stacks the same height.
Output a blank line between each set.
Sample Input
6
5 2 4 1 7 5
0
Sample Output
5
题目分析:
哈哈 ,又是一个水题, 貌似这段时间一直在水题. 刚开始看的时候还以为是 DP , 看懂题目后知道这只是一个简单的小学数学题,呵呵. 如果求移动步数还复杂点, 但题目要求的是 最小搬动次数, 那么只要求出平均值, 大于平均值的墙就是要移动的墙 ,累加就可以了.
代码如下:
MiYu原创, 转帖请注明 : 转载自 ______________白白の屋
#include <iostream>
#include <string>
using namespace std;
int main ()
{
int N;
int f = 0;
while ( cin >> N, N )
{
if ( f )
{
cout << endl;
}
f = 1;
int num[N+1];
num[N] = 0;
for ( int i = 0; i != N; ++ i )
{
cin >> num[i];
num[N] += num[i];
}
num[N] /= N;
int nCount = 0;
for ( int i = 0; i != N; ++ i )
{
if ( num[i] > num[N] )
{
nCount += num[i] - num[N];
}
}
cout << nCount << endl;
}
return 0;
}