《ACM程序设计》书中题目―K

    xiaoxiao2021-03-25  43

    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.

    1、题意:该题意思为将不同的高度的墙通过移动砖头,最终变为一样高,也就是将不同大小的数通过加减变为他们的平均数。

    2、思路:需要求出他们的平均数,然后用平均数依次减去比它小的数字,然后再求和。

    3、代码:

    #include<iostream> using namespace std; int main() { int i,t,n,p,y=0; int a[100]={0}; while(cin>>n) { if(n==0) break; y++; for(i=0,t=0;i<n;i++) { cin>>a[i]; t+=a[i]; } t=t/n; for(i=0,p=0;i<n;i++) { if(a[i]<t) p+=t-a[i]; else continue; } cout<<"Set #"<<y<<endl; cout<<"The minimum number of moves is "<<p<<"."<<endl; cout<<endl; } return 0; }4、总结:这道题挺简单的,我却因为输出时漏掉一个点还有空行以及打错字母而几次未通过,以后要更加注意细节,不能潦草。

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