zoukankan      html  css  js  c++  java
  • [POJ1477]Box of Bricks


    Time Limit: 1000MS   Memory Limit: 10000K
    Total Submissions: 19503   Accepted: 7871

    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

    6
    5 2 4 1 7 5
    0

    Sample Output

    Set #1
    The minimum number of moves is 5.
    

    Source

    THINKING

      贪心,先计算平均值,再扫描数组求出差值,注意句末的句号和组数之间的空行!

    var a:array[1..10000] of longint;
        n,i,sum,summ,t:longint;
    begin
        summ:=1;
        while true do
            begin
                fillchar(a,sizeof(a),0);
                sum:=0;
                t:=0;
                readln(n);
                if n=0 then halt;
                for i:=1 to n do
                    begin
                        read(a[i]);
                        inc(sum,a[i]);
                    end;
                sum:=sum div n;
                for i:=1 to n do
                    t:=abs(sum-a[i])+t;
                writeln('Set #',summ);
                writeln('The minimum number of moves is ',t div 2,'.');
                writeln;
                inc(summ);
            end;
    end.
    View Code
  • 相关阅读:
    sql中别名加as和不加as的区别
    easyui 扩展 datagrid 数据网格视图
    asp.net 版本一键升级,后台直接调用升级脚本
    SQLserver 还原数据库报“指定转换无效”的错的解决方案
    sql视图显示数据不对应
    django channels 实现实时通讯,主动推送
    django orm信号机制 + apschedule 定时任务
    django 使用原始SQL语句方式
    Django model 常用查询(搬运来备份的)
    Python3+ Django2.7开发web排坑记006 apache+wsgi部署应用
  • 原文地址:https://www.cnblogs.com/yangqingli/p/4889801.html
Copyright © 2011-2022 走看看