地址:http://codeforces.com/contest/801/problem/C

题目:

C. Voltage Keepsake
time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You have n devices that you want to use simultaneously.

The i-th device uses ai units of power per second. This usage is continuous. That is, in λ seconds, the device will use λ·ai units of power. The i-th device currently has bi units of power stored. All devices can store an arbitrary amount of power.

You have a single charger that can plug to any single device. The charger will add p units of power per second to a device. This charging is continuous. That is, if you plug in a device for λ seconds, it will gain λ·p units of power. You can switch which device is charging at any arbitrary unit of time (including real numbers), and the time it takes to switch is negligible.

You are wondering, what is the maximum amount of time you can use the devices until one of them hits 0 units of power.

If you can use the devices indefinitely, print -1. Otherwise, print the maximum amount of time before any one device hits 0 power.

Input

The first line contains two integers, n and p (1 ≤ n ≤ 100 000, 1 ≤ p ≤ 109) — the number of devices and the power of the charger.

This is followed by n lines which contain two integers each. Line i contains the integers ai and bi (1 ≤ ai, bi ≤ 100 000) — the power of the device and the amount of power stored in the device in the beginning.

Output

If you can use the devices indefinitely, print -1. Otherwise, print the maximum amount of time before any one device hits 0 power.

Your answer will be considered correct if its absolute or relative error does not exceed 10 - 4.

Namely, let's assume that your answer is a and the answer of the jury is b. The checker program will consider your answer correct if .

Examples
input
2 1
2 2
2 1000
output
2.0000000000
input
1 100
1 1
output
-1
input
3 5
4 3
5 2
6 1
output
0.5000000000
Note

In sample test 1, you can charge the first device for the entire time until it hits zero power. The second device has enough power to last this time without being charged.

In sample test 2, you can use the device indefinitely.

In sample test 3, we can charge the third device for 2 / 5 of a second, then switch to charge the second device for a 1 / 10 of a second.

思路:二分时间加check

  ps1:二分这种小数选用规定二分100次(2^100)或者更多即可,尽量不要用while(fabs(r-l)>eps)

  ps2:二分上线r最少应该是1e10,不然会被hack(比赛结束后看其他人讨论时才知道这个hack点的,一阵后怕,还好我刚好写的是1e10)

 #include <bits/stdc++.h>

 using namespace std;

 #define MP make_pair
#define PB push_back
typedef long long LL;
typedef pair<int,int> PII;
const double eps=1e-;
const double pi=acos(-1.0);
const int K=1e6+;
const int mod=1e9+; int n,p,a[K],b[K];
double ans;
bool check(double x)
{
double sum=;
for(int i=;i<=n;i++)
{
sum+=max(a[i]*x-b[i],0.0);
if(sum>=x*p+1e-)
return ;
}
return ;
}
int main(void)
{
cin>>n>>p;
for(int i=;i<=n;i++)
scanf("%d%d",a+i,b+i);
double l=,r=1e10;
for(int i=;i<=;i++)
{
double mid=(l+r)/2.0;
if(check(mid))
ans=mid,l=mid;
else
r=mid;
}
if(fabs(ans-1e10)<=1e-)
printf("-1\n");
else
printf("%.8f\n",ans);
return ;
}

Codeforces Round #409 (rated, Div. 2, based on VK Cup 2017 Round 2) C Voltage Keepsake的更多相关文章

  1. Codeforces Round #409 (rated, Div. 2, based on VK Cup 2017 Round 2)(A.思维题,B.思维题)

    A. Vicious Keyboard time limit per test:2 seconds memory limit per test:256 megabytes input:standard ...

  2. Codeforces Round #409 (rated, Div. 2, based on VK Cup 2017 Round 2) D. Volatile Kite

    地址:http://codeforces.com/contest/801/problem/D 题目: D. Volatile Kite time limit per test 2 seconds me ...

  3. Codeforces Round #409 (rated, Div. 2, based on VK Cup 2017 Round 2) 题解【ABCDE】

    A. Vicious Keyboard 题意:给你一个字符串,里面只会包含VK,这两种字符,然后你可以改变一个字符,你要求VK这个字串出现的次数最多. 题解:数据范围很小,暴力枚举改变哪个字符,然后c ...

  4. Codeforces Round #409 (rated, Div. 2, based on VK Cup 2017 Round 2)

    A 每次可以换一个或不换,暴力枚举位置即可 B 模拟 C 二分答案.. 边界可以优化r=totb/(tota-p),二分可以直接(r-l>=EPS,EPS不要太小,合适就好),也可以直接限定二分 ...

  5. Codeforces Round #409 (rated, Div. 2, based on VK Cup 2017 Round 2) A B C D 暴力 水 二分 几何

    A. Vicious Keyboard time limit per test 2 seconds memory limit per test 256 megabytes input standard ...

  6. Codeforces Round #405 (rated, Div. 2, based on VK Cup 2017 Round 1) 菜鸡只会ABC!

    Codeforces Round #405 (rated, Div. 2, based on VK Cup 2017 Round 1) 全场题解 菜鸡只会A+B+C,呈上题解: A. Bear and ...

  7. Codeforces Round #405 (rated, Div. 2, based on VK Cup 2017 Round 1) C. Bear and Different Names 贪心

    C. Bear and Different Names 题目连接: http://codeforces.com/contest/791/problem/C Description In the arm ...

  8. Codeforces Round #405 (rated, Div. 2, based on VK Cup 2017 Round 1) B - Bear and Friendship Condition 水题

    B. Bear and Friendship Condition 题目连接: http://codeforces.com/contest/791/problem/B Description Bear ...

  9. 【树形dp】Codeforces Round #405 (rated, Div. 1, based on VK Cup 2017 Round 1) B. Bear and Tree Jumps

    我们要统计的答案是sigma([L/K]),L为路径的长度,中括号表示上取整. [L/K]化简一下就是(L+f(L,K))/K,f(L,K)表示长度为L的路径要想达到K的整数倍,还要加上多少. 于是, ...

随机推荐

  1. 带清空按钮TextBox的实现(WPF)

    本博文针对人群:WPF新手.博文内容:通过Style制定包含清空Button的TextBox样式模板,通过在Style中引入自定义类的附加属性完成对TextBox的内容清空. <span sty ...

  2. jQuery动态星级评分效果实现方法

    本文实例讲述了jQuery动态星级评分效果实现方法.分享给大家供大家参考.具体如下: 这里的jQuery星级评分代码,是大家都很喜欢的功能,目前广泛应用,本星级评分加入了动画效果,注意,如果要真正实现 ...

  3. 简单脱壳教程笔记(8)---手脱EZIP壳

    本笔记是针对ximo早期发的脱壳基础视频教程,整理的笔记.本笔记用到的工具下载地址: http://download.csdn.net/detail/obuyiseng/9466056 EZIP壳 : ...

  4. HDU2586.How far away ?——近期公共祖先(离线Tarjan)

    http://acm.hdu.edu.cn/showproblem.php?pid=2586 给定一棵带权有根树,对于m个查询(u,v),求得u到v之间的最短距离 那么仅仅要求得LCA(u,v),di ...

  5. Spring_day04--整合struts2和spring框架

    整合struts2和spring框架 1 把struts2的action交给spring管理 2 实现过程 第一步 导入struts2和spring的jar包 (1)导入用于整合的jar包 第二步 创 ...

  6. Fragment之间传数据

    1.用bundle存Bundle bundle = new Bundle();bundle.putString("cid1", classList.get(i).getId()); ...

  7. 工作表(Worksheet)基本操作应用示例

    在编写代码时,经常要引用工作表的名字.知道工作表在工作簿中的位置.增加工作表.删除工作表.复制工作表.移动工作表.重命名工作表,等等.下面介绍与此有关及相关的一些属性和方法示例. [示例04-01]增 ...

  8. JZOJ.5289【NOIP2017模拟8.17】偷笑

    Description berber走进机房,边敲门边喊:“我是哔哔”CRAZY转过头:“我警告你,哔哔刚刚来过!”“呵呵呵呵……”这时,哔哔站了起来,环顾四周:“你们笑什么?……”巧了,发出笑声的人 ...

  9. [SCOI2010]字符串

    思路: 设1为向(1,1)方向走,0为向(1,-1)方向走.那么题意可转化为从(0,0)走到(n+m,n-m)且不能跨过y=0的方案数.总方案数C(n+m,n),然后要减去不合法的即线路通过y=-1的 ...

  10. 前端性能优化-减少http请求,dns预解析,减少repaint和reflow

    前端性能优化方法: 一 . 减少http请求 (1)通过合并图片,减少请求,俗称css sprites(css精灵)css sprites (2)lazyload懒加载,在需要的时候再加载 1.定义: ...