POJ 3122 pie 二分
来源:互联网 发布:java数据结构与算法题 编辑:程序博客网 时间:2024/05/20 09:09
Description
My birthday is coming up and traditionally I"m serving pie. Not just one pie, no, I have a number N of them, of various tastes and of various sizes. F of my friends are coming to my party and each of them gets a piece of pie. This should be one piece of one pie, not several small pieces since that looks messy. This piece can be one whole pie though.My friends are very annoying and if one of them gets a bigger piece than the others, they start complaining. Therefore all of them should get equally sized (but not necessarily equally shaped) pieces, even if this leads to some pie getting spoiled (which is better than spoiling the party). Of course, I want a piece of pie for myself too, and that piece should also be of the same size.
What is the largest possible piece size all of us can get? All the pies are cylindrical in shape and they all have the same height 1, but the radii of the pies can be different.
Input
One line with a positive integer: the number of test cases. Then for each test case:One line with two integers N and F with 1 ≤ N, F ≤ 10 000: the number of pies and the number of friends.One line with N integers ri with 1 ≤ ri ≤ 10 000: the radii of the pies.Output
For each test case, output one line with the largest possible volume V such that me and my friends can all get a pie piece of size V. The answer should be given as a floating point number with an absolute error of at most 10?3.Sample Input
33 34 3 31 24510 51 4 2 3 4 5 6 5 4 2
Sample Output
25.13273.141650.2655
有n块蛋糕,分给f+1个人,但是每个人必须是整块(不能来自不同的两块蛋糕)。
思路:二分法,答案在总体积除以人数(sum/(f+1))到0之间寻找,满足条件切精确度足够则范围值。
#include <iostream>#include <string.h>#include <stdio.h>#include <math.h>#include <map>#include<queue>#define pi 3.14159265359using namespace std;int main(){ int t; cin>>t; while(t--) { int n,f; double v[10005],max=0,mid=-10000,min=0; cin>>n>>f; for(int i=0;i<n;i++) { cin>>v[i]; v[i]=v[i]*v[i]*pi; max+=v[i]; } max=max/(f+1); while(fabs(min-max)>0.0001) { int sum=0,flag=1; mid=(max+min)/2; for(int i=0;i<n;i++) { sum+=(int)floor(v[i]/mid); if(sum>=f+1) { min=mid; flag=0; break; } } if(flag) max=mid; } printf("%.4lf\n",min); }}
- POJ 3122 Pie 二分
- POJ 3122 Pie 二分
- poj 3122 Pie (二分)
- Pie - POJ 3122 二分
- POJ 3122 Pie(二分)
- POJ 3122pie(二分)
- POJ 3122 pie 二分
- [poj 3122] Pie 二分
- POJ Pie 3122 (二分)
- POJ 3122 - Pie(二分)
- poj 3122 pie(二分搜索)
- poj 3122 Pie 二分答案
- POJ 3122 Pie【二分答案】
- poj 3122&&hdu1969 Pie(二分)
- poj 3122 Pie (二分查找)
- POJ 3122-Pie(二分+精度)
- 【POJ 3122】 Pie (二分+贪心)
- Pie(POJ--3122【二分查找】
- Java自定义比较器实现中文排序
- Data Structures And Problem Solving Using Java (Fourth Edition)中译版(Java 修饰词与可见性)
- HTML窗体指南
- Redis3.0与Jedis2.7.2 客户端与Spring整合
- SPARK排序算法,使用Scala开发 二次排序 自定义KEY值,相比JAVA的罗嗦,Scala优雅简洁!!!
- POJ 3122 pie 二分
- MyBatis自动生成Entity、Dao、Mapping
- Git学习笔记之分支
- ionic css布局之bar页眉页脚定义
- C++第1次实验(提高班)——复习
- BSTR、char*和CString转换
- 二级指针
- LNMP环境搭建
- eclipse新建安卓项目点击finish后窗口无法关闭