zoukankan      html  css  js  c++  java
  • HDU 3420 -- Bus Fair ACM

    Bus Fair

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 600    Accepted Submission(s): 293

    Problem Description

    You are now in Foolish Land. Once moving in Foolish Land you found that there is a strange Bus fair system. The fair of moving one kilometer by bus in that country is one coin. If you want to go to X km and your friend wants to go to Y km then you can buy a ticket of X+Y coins (you are also allowed to buy two or more tickets for you two).    Now as a programmer, you want to show your creativity in buying tickets! Suppose, your friend wants to go 1 km and you want to go 2 km. Then it’s enough for you to buy a 2coin ticket! Because both of you are valid passengers before crossing the first km. and when your bus cross the first km your friend gets down from the bus. So you have the ticket of 2km! And you can safely reach to your destination, 2km using that ticket.    Now, you have a large group of friends and they want to reach to different distance. You think that you are smart enough that you can buy tickets that should manage all to reach their destination spending the minimum amount of coins. Then tell us how much we should at least pay to reach our destination.

    Input

    There are multiple test cases. Each case start with a integer n, the total number of people in that group. 0<=n<=1000. Then comes n integers, each of them stands for a distance one of the men of the group wants to go to. You can assume that the distance a man wants to go is always less than 10000.

    Output

    Your program should print a single integer for a single case, the minimum amount of coins the group should spend to reach to the destination of all the members of that group.

    Sample Input

    2
    1
    2
    2
    2
    3

    Sample Output

    2
    4

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3420
     1 #include<stdio.h>
     2 #include<stdlib.h>
     3 int cmp(const void* a,const void* b)
     4 {
     5     return *(int *)a - *(int *)b;
     6 }
     7 int main()
     8 {
     9     int n,tmp,sum,i;
    10     int num[10000];
    11     while(~scanf("%d",&n))
    12     {
    13         for(i=0;i<n;i++)
    14             scanf("%d",&num[i]);
    15         sum=0;
    16         qsort(num,n,sizeof(num[0]),cmp);
    17         for(i=0;i<n;i++)
    18         {
    19             tmp=num[i]*(n-i);
    20             if(tmp>sum)
    21                 sum=tmp;
    22         }
    23         printf("%d
    ",sum);
    24     }
    25     return 0;
    26 }


     

  • 相关阅读:
    项目管理【53】 | 项目风险管理-规划风险应对
    Learning a Continuous Representation of 3D Molecular Structures with Deep Generative Models
    转:DenseNet
    转:期刊投稿中的简写(ADM,AE,EIC等)与流程状态解读
    论文中如何写算法伪代码
    氨基酸,多肽,蛋白质等
    pytorch查看全连接层的权重和梯度
    AI制药文章
    long-tail datasets
    转:Focal Loss理解
  • 原文地址:https://www.cnblogs.com/yeshadow937/p/3904209.html
Copyright © 2011-2022 走看看