HDOJ(HDU) 2304 Electrical Outlets(求和、、)

举报
谙忆 发表于 2021/05/28 07:16:09 2021/05/28
【摘要】 Problem Description Roy has just moved into a new apartment. Well, actually the apartment itself is not very new, even dating back to the days before people had electricity in their hou...

Problem Description
Roy has just moved into a new apartment. Well, actually the apartment itself is not very new, even dating back to the days before people had electricity in their houses. Because of this, Roy’s apartment has only one single wall outlet, so Roy can only power one of his electrical appliances at a time.
Roy likes to watch TV as he works on his computer, and to listen to his HiFi system (on high volume) while he vacuums, so using just the single outlet is not an option. Actually, he wants to have all his appliances connected to a powered outlet, all the time. The answer, of course, is power strips, and Roy has some old ones that he used in his old apartment. However, that apartment had many more wall outlets, so he is not sure whether his power strips will provide him with enough outlets now.
Your task is to help Roy compute how many appliances he can provide with electricity, given a set of power strips. Note that without any power strips, Roy can power one single appliance through the wall outlet. Also, remember that a power strip has to be powered itself to be of any use.

Input
Input will start with a single integer 1 <= N <= 20, indicating the number of test cases to follow. Then follow N lines, each describing a test case. Each test case starts with an integer 1 <= K <= 10, indicating the number of power strips in the test case. Then follow, on the same line, K integers separated by single spaces, O1 O2 … OK, where 2 <= Oi <= 10, indicating the number of outlets in each power strip.

Output
Output one line per test case, with the maximum number of appliances that can be powered.

Sample Input
3
3 2 3 4
10 4 4 4 4 4 4 4 4 4 4
4 10 10 10 10

Sample Output
7
31
37

这个是关于排插的问题,就是给你m个排插,每个排插上有x个孔,最开始只有一个墙壁能插一个排插,问能插多少个电器,一个电器要占一个孔。

也就是把这些排插给串起来,串起来的排插也需要占一个孔,然后因为最后一个排插不用被排插占一个孔,所以不用减一,
所以答案就是:m个排插的孔全部加起来,再减去(m-1);

import java.util.Scanner;

public class Main{ public static void main(String[] args) { Scanner sc = new Scanner(System.in); int t =sc.nextInt(); while(t-->0){ int n =sc.nextInt(); int sum=0; int a; for(int i=0;i<n;i++){ a=sc.nextInt(); sum+=a; } sum=sum-n+1; System.out.println(sum); } }
}

  
 
  • 1
  • 2
  • 3
  • 4
  • 5
  • 6
  • 7
  • 8
  • 9
  • 10
  • 11
  • 12
  • 13
  • 14
  • 15
  • 16
  • 17
  • 18
  • 19
  • 20
  • 21

文章来源: chenhx.blog.csdn.net,作者:谙忆,版权归原作者所有,如需转载,请联系作者。

原文链接:chenhx.blog.csdn.net/article/details/51334364

【版权声明】本文为华为云社区用户转载文章,如果您发现本社区中有涉嫌抄袭的内容,欢迎发送邮件进行举报,并提供相关证据,一经查实,本社区将立刻删除涉嫌侵权内容,举报邮箱: cloudbbs@huaweicloud.com
  • 点赞
  • 收藏
  • 关注作者

评论(0

0/1000
抱歉,系统识别当前为高风险访问,暂不支持该操作

全部回复

上滑加载中

设置昵称

在此一键设置昵称,即可参与社区互动!

*长度不超过10个汉字或20个英文字符,设置后3个月内不可修改。

*长度不超过10个汉字或20个英文字符,设置后3个月内不可修改。