[POJ1477]Box of Bricks

时间：2015-10-18 18:33:07 阅读：411 评论：0 收藏：0 [点我收藏+]

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?
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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

Southwestern European Regional Contest 1997

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

[POJ1477]Box of Bricks

原文：http://www.cnblogs.com/yangqingli/p/4889801.html

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