三分显然,要注意EPS必须设成1e-6,设得再小一点都会TLE……坑炸了

#include<cstdio>
#include<algorithm>
#include<cmath>
using namespace std;
#define EPS 0.000001
int n,x[60010],v[60010];
double calc(double p)
{
double res=0;
for(int i=1;i<=n;++i)
res=max(res,fabs(p-(double)x[i])/(double)v[i]);
return res;
}
int main()
{
// freopen("b.in","r",stdin);
scanf("%d",&n);
for(int i=1;i<=n;++i)
scanf("%d",&x[i]);
for(int i=1;i<=n;++i)
scanf("%d",&v[i]);
double l=1.0,r=1000000000.0;
while(r-l>EPS)
{
double m1=l+(r-l)/3.0;
double m2=r-(r-l)/3.0;
double t1=calc(m1),t2=calc(m2);
if(t1>t2)
l=m1;
else
r=m2;
}
printf("%.12lf\n",calc(l));
return 0;
}

【三分】Codeforces Round #403 (Div. 2, based on Technocup 2017 Finals) B. The Meeting Place Cannot Be Changed的更多相关文章

  1. Codeforces Round #403 (Div. 2, based on Technocup 2017 Finals) B. The Meeting Place Cannot Be Changed

    地址:http://codeforces.com/contest/782/problem/B 题目: B. The Meeting Place Cannot Be Changed time limit ...

  2. Codeforces Round #403 (Div. 2, based on Technocup 2017 Finals)

    Codeforces Round #403 (Div. 2, based on Technocup 2017 Finals) 说一点东西: 昨天晚上$9:05$开始太不好了,我在学校学校$9:40$放 ...

  3. Codeforces Round #403 (Div. 2, based on Technocup 2017 Finals)A模拟 B三分 C dfs D map

    A. Andryusha and Socks time limit per test 2 seconds memory limit per test 256 megabytes input stand ...

  4. 树的性质和dfs的性质 Codeforces Round #403 (Div. 2, based on Technocup 2017 Finals) E

    http://codeforces.com/contest/782/problem/E 题目大意: 有n个节点,m条边,k个人,k个人中每个人都可以从任意起点开始走(2*n)/k步,且这个步数是向上取 ...

  5. 2-sat Codeforces Round #403 (Div. 2, based on Technocup 2017 Finals) D

    http://codeforces.com/contest/782/problem/D 题意: 每个队有两种队名,问有没有满足以下两个条件的命名方法: ①任意两个队的名字不相同. ②若某个队 A 选用 ...

  6. Codeforces Round #403 (Div. 2, based on Technocup 2017 Finals) E Underground Lab

    地址:http://codeforces.com/contest/782/problem/E 题目: E. Underground Lab time limit per test 1 second m ...

  7. Codeforces Round #403 (Div. 2, based on Technocup 2017 Finals) D. Innokenty and a Football League

    地址:http://codeforces.com/contest/782/problem/D 题目: D. Innokenty and a Football League time limit per ...

  8. Codeforces Round #403 (Div. 2, based on Technocup 2017 Finals) C Andryusha and Colored Balloons

    地址:http://codeforces.com/contest/782/problem/C 题目: C. Andryusha and Colored Balloons time limit per ...

  9. Codeforces Round #403 (Div. 2, based on Technocup 2017 Finals) A. Andryusha and Socks

    地址:http://codeforces.com/contest/782/problem/A 题目: A. Andryusha and Socks time limit per test 2 seco ...

随机推荐

  1. JSONP以及Spring对象MappingJacksonValue的使用方式

    什么是JSONP?,以及Spring对象MappingJacksonValue的使用方式 原文: https://blog.csdn.net/weixin_38111957/article/detai ...

  2. HDU 5671 矩阵

    Matrix Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Su ...

  3. AnnotationConfigApplicationContext.的用法的核心代码

    public static void main(String[] args) {ApplicationContext ctx = new AnnotationConfigApplicationCont ...

  4. docker公司测试环境搭建总结

    1.防火墙转发规则: [root@docker ~]# firewall-cmd --list-allpublic (active) target: default icmp-block-invers ...

  5. offset--BUG

    offsetWidth所获取的宽度并不是div的实际宽度,它包括div的width.border等. 在JS函数中,可以通过obj.style.width来获取div的实际宽度,但是这种方式style ...

  6. jquery学习总计

    1,jquery的基础语法 $(selector).action(); 选择器(selector)查询和查找html元素,action()执行对函数的操作. 2.选择器 id,类,类型,属性,属性值等 ...

  7. httpFS访问

    编辑文件httpfs-env.sh 执行sbin/httpfs.sh 执行命令curl -i "http://192.168.1.213:14000/webhdfs/v1?user.name ...

  8. bzoj1036: [ZJOI2008]树的统计Count link-cut-tree版

    题目传送门 这 算是link-cut-tree裸题啊 不过以前好像没有写过单点修改.............. #include<cstdio> #include<cstring&g ...

  9. float/文档流

    float : left | right | none | inherit; 文档流是文档中可显示对象在排列时所占用的位置. 浮动的定义: 使元素脱离文档流,按照指定方向发生移动,遇到父级边界或者相邻 ...

  10. 流程控制 while循环 运算符

    具体知识戳这里 可变数据类型:在id不变的情况下,数据类型内部的元素(value)可以改变 如:列表,字典 不可变类型:value改变,id也跟的改变 如:数字.字符.布尔类型 运算符 #算数运算符# ...