Cable master(好题,二分)
Cable master
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2473 Accepted Submission(s): 922
To buy network cables, the Judging Committee has contacted a local network solutions provider with a request to sell for them a specified number of cables with equal lengths. The Judging Committee wants the cables to be as long as possible to sit contestants as far from each other as possible.
The Cable Master of the company was assigned to the task. He knows the length of each cable in the stock up to a centimeter, and he can cut them with a centimeter precision being told the length of the pieces he must cut. However, this time, the length is not known and the Cable Master is completely puzzled.
You are to help the Cable Master, by writing a program that will determine the maximal possible length of a cable piece that can be cut from the cables in the stock, to get the specified number of pieces.
The input is ended by line containing two 0's.
If it is not possible to cut the requested number of pieces each one being at least one centimeter long, then the output must contain the single number "0.00" (without quotes).
#include<stdio.h>
double m[];
int N;double sum,K;
double search(){int temp=;
double l=,r=sum/K,mid;
while(r-l>1e-){temp=;
mid=(l+r)/;
for(int i=;i<N;++i)temp+=(int)(m[i]/mid);//temp代表能裁剪成的数量。。。
if(temp>=K)l=mid;
else r=mid;
}
return mid;
}
int main(){
while(scanf("%d%lf",&N,&K),N||K){sum=;
for(int i=;i<N;++i)scanf("%lf",&m[i]),sum+=m[i];
printf("%.2lf\n",search());
}
return ;
}
poj ac:
#include<stdio.h>
double m[];
int N;double sum,K;
double search(){int temp=;
double l=,r=sum/K,mid;
while(r-l>1e-){temp=;
mid=(l+r)/;
for(int i=;i<N;++i)temp+=(int)(m[i]/mid);
if(temp>=K)l=mid;
else r=mid;
}
return r;
}
int main(){double d;double t;
while(~scanf("%d%lf",&N,&K)){sum=;
for(int i=;i<N;++i)scanf("%lf",&m[i]),sum+=m[i];
d=search();
/// t=100*d;printf("%d\n",t);
printf("%.2f\n",(int)(d*)/100.0);
}
return ;
}
Cable master(好题,二分)的更多相关文章
- poj 1064 Cable master【浮点型二分查找】
Cable master Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 29554 Accepted: 6247 Des ...
- Cable master poj1064(二分)
http://poj.org/problem?id=1064 题意:共有n段绳子,要求总共被分为k段.问在符合题意的前提下,每段长最大是多少? #include <iostream> #i ...
- Cable master(二分题 注意精度)
Cable master Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 26596 Accepted: 5673 Des ...
- POJ 1064 Cable master(二分查找+精度)(神坑题)
POJ 1064 Cable master 一开始把 int C(double x) 里面写成了 int C(int x) ,莫名奇妙竟然过了样例,交了以后直接就wa. 后来发现又把二分查找的判断条 ...
- [ACM] poj 1064 Cable master (二分查找)
Cable master Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 21071 Accepted: 4542 Des ...
- (poj)1064 Cable master 二分+精度
题目链接:http://poj.org/problem?id=1064 Description Inhabitants of the Wonderland have decided to hold a ...
- poj 1064 Cable master 判断一个解是否可行 浮点数二分
poj 1064 Cable master 判断一个解是否可行 浮点数二分 题目链接: http://poj.org/problem?id=1064 思路: 二分答案,floor函数防止四舍五入 代码 ...
- POJ 1064 Cable master (二分查找)
题目链接 Description Inhabitants of the Wonderland have decided to hold a regional programming contest. ...
- Cable master(二分-求可行解)
Inhabitants of the Wonderland have decided to hold a regional programming contest. The Judging Commi ...
随机推荐
- android-sdk-windows版本号下载
Android SDK 4.0.3 开发环境配置及执行 近期又装了一次最新版本号的ADK环境 眼下最新版是Android SDK 4.0.3 本文的插图和文本尽管是Android2.2的 步骤都是一样 ...
- 为项目编写Readme.MD文件
了解一个项目,恐怕首先都是通过其Readme文件了解信息.如果你以为Readme文件都是随便写写的那你就错了.github,oschina git gitcafe的代码托管平台上的项目的Readme. ...
- JS实现下拉框选中不同的项,对应显示不同的信息
实现的效果如下图: 页面代码 下拉框: <select id="select3" name="select3" onchange="showli ...
- Jquery根据字段内容设置字段宽度
来博客园很久了,初次写文章,新手,请大牛见谅! 前段时间遇到的问题,通过gridview后台动态生成table,列和行数量未知,要求根据每个单元格内容的多少,设置宽度,每一列选择本列最大的宽度. ta ...
- oracle 两个时间相减
oracle 两个时间相减默认的是天数 oracle 两个时间相减默认的是天数*24 为相差的小时数 oracle 两个时间相减默认的是天数*24*60 为相差的分钟数 oracle 两个时间相减默认 ...
- 驱动编程思想之初体验 --------------- 嵌入式linux驱动开发之点亮LED
这节我们就开始开始进行实战啦!这里顺便说一下啊,出来做开发的基础很重要啊,基础不好,迟早是要恶补的.个人深刻觉得像这种嵌入式的开发对C语言和微机接口与原理是非常依赖的,必须要有深厚的基础才能hold的 ...
- A Typical Homework(学生信息管理系统)
A Typical Homework(a.k.a Shi Xiong Bang Bang Mang) Hi, I am an undergraduate student in institute of ...
- TCP的状态
在TCP层,有个FLAGS字段,这个字段有以下几个标识:SYN, FIN, ACK, PSH, RST, URG. 其中,对于我们日常的分析有用的就是前面的五个字段. 它们的含义是: SYN表示建立连 ...
- (原+转)error C2872: “ACCESS_MASK”: 不明确的符号 C:\Program Files (x86)\Windows Kits\8.1\Include\um\ktmw32.h 125
转载请注明出处: http://www.cnblogs.com/darkknightzh/p/5500795.html 参考网址: http://www.lxway.com/88515256.htm ...
- YUI Array 之some(检测|any)
YUI原码 YUI someYArray.some = Lang._isNative(Native.some) ? function (array, fn, thisObj) { return Nat ...