U-Box of Bricks

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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 #1

The minimum number of moves is 5.

题目描述

移动最小的积木,使每排积木的高相等,还要求每排必须相邻,求最小的移动积木数

解题思路

求移动最少的数目使所有的数目相等,我们可以用贪心的思想来想这个题,先求出平均的高度,不管比这个高度低的,将比它高的所有的木块取出,就能得出结果

源代码

#include<iostream>#include<cstdio>#include<algorithm>#include<cstring>using namespace std;int a[1000];int main(){        int n,sum,x,y,f=0;        while(cin>>n&&n)        {                f++;                int i;                sum=0;                y=0;                for(i=0;i<n;i++)                {                        cin>>a[i];                        sum+=a[i];                }                x=sum/n;                for(i=0;i<n;i++)                {                        if(a[i]>x)                                y+=a[i]-x;                }                cout<<"Set #"<<f<<endl;                cout<<"The minimum number of moves is "<<y<<"."<<endl<<endl;        }        return 0;}

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