zoukankan      html  css  js  c++  java
  • UVA

    最多时间就是每仅仅蚂蚁选择最久的爬行方式


    最少时间就是每仅仅蚂蚁选择最快地爬行方式

    #include<iostream>
    #include<map>
    #include<string>
    #include<cstring>
    #include<cstdio>
    #include<cstdlib>
    #include<cmath>
    #include<queue>
    #include<vector>
    #include<algorithm>
    using namespace std;
    int main()
    {
    	int a,b,i,T,len,n,t;
    	cin>>T;
    	while(T--)
    	{
    		cin>>len>>n;
    		a=b=0;
    		while(n--)
    		{
    			cin>>t;
    			a=max(a,min(len-t,t));
    			b=max(b,max(len-t,t));
    		}
    		cout<<a<<" "<<b<<endl;
    	}
    	return 0;
    }


    Time Limit:3000MS   Memory Limit:Unknown   64bit IO Format:%lld & %llu

    Status

    Description

    Download as PDF

    Problem B: Ants

    An army of ants walk on a horizontal pole of length l cm, each with a constant speed of 1 cm/s. When a walking ant reaches an end of the pole, it immediatelly falls off it. When two ants meet they turn back and start walking in opposite directions. We know the original positions of ants on the pole, unfortunately, we do not know the directions in which the ants are walking. Your task is to compute the earliest and the latest possible times needed for all ants to fall off the pole.

    The first line of input contains one integer giving the number of cases that follow. The data for each case start with two integer numbers: the length of the pole (in cm) andn, the number of ants residing on the pole. These two numbers are followed byn integers giving the position of each ant on the pole as the distance measured from the left end of the pole, in no particular order. All input integers are not bigger than 1000000 and they are separated by whitespace.

    For each case of input, output two numbers separated by a single space. The first number is the earliest possible time when all ants fall off the pole (if the directions of their walks are chosen appropriately) and the second number is the latest possible such time.

    Sample input

     
    2
    10 3
    2 6 7
    214 7
    11 12 7 13 176 23 191
    

    Output for sample input

    4 8
    38 207
    

    Piotr Rudnicki

    Source

    Root :: Competitive Programming 3: The New Lower Bound of Programming Contests (Steven & Felix Halim) :: Problem Solving Paradigms :: Greedy ::Non Classical, Usually Harder
    Root :: AOAPC I: Beginning Algorithm Contests (Rujia Liu) :: Volume 4. Algorithm Design

    Root :: Competitive Programming 2: This increases the lower bound of Programming Contests. Again (Steven & Felix Halim) :: Problem Solving Paradigms ::Greedy - Standard
    Root :: Prominent Problemsetters :: Piotr Rudnicki

  • 相关阅读:
    JS使用Enter事件将输入的字符倒叙输出
    博客园 页面定制CSS代码
    C#将DataTable数据导出CSV文件
    js时间格式转换
    将十进制数据转换为二进制AND将二进制数据转换为十进制!
    ref和out的使用与区别
    形参和实参的区别
    为啥Underlay才是容器网络的最佳落地选择
    你问我答:容器篇(1)
    切忌一步到位,谈谈DevOps实施落地
  • 原文地址:https://www.cnblogs.com/mengfanrong/p/5092620.html
Copyright © 2011-2022 走看看