【hdu1969】Pie——二分
2016-03-13 23:48
399 查看
题目:
Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7795 Accepted Submission(s): 2889
Problem 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
Sample Output
Source
NWERC2006
描述:给N个等高圆柱体的半径,求把他们分给F+1个人的最大体积
题解:二分裸题,但是要注意精度,1e-6正好A,1e-8蜜汁WA
代码:
Pie
Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 7795 Accepted Submission(s): 2889
Problem 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
3 3 3 4 3 3 1 24 5 10 5 1 4 2 3 4 5 6 5 4 2
Sample Output
25.1327 3.1416 50.2655
Source
NWERC2006
描述:给N个等高圆柱体的半径,求把他们分给F+1个人的最大体积
题解:二分裸题,但是要注意精度,1e-6正好A,1e-8蜜汁WA
代码:
#include <cstdio> #include <algorithm> #include <cstring> #include <cmath> using namespace std; const double eps = 1e-6; const int maxn = 10005; const double pi = acos(-1.0); int r[maxn]; int main() { //freopen("input.txt", "r", stdin); int T; scanf("%d", &T); while (T--) { int N, F; scanf("%d%d", &N, &F); for (int i = 0; i < N; i++) { scanf("%d", &r[i]); } sort(r, r + N); double l = 0.0; double r1 = pi*r[N - 1] * r[N - 1]; while (r1 - l > eps) { double mid = (l + r1) / 2; int temp = 0; for (int i = 0; i < N; i++) temp += floor(pi*r[i] * r[i] / mid); if (temp >= F + 1) l = mid; else r1 = mid; } printf("%.4lf\n", l); } return 0; }
相关文章推荐
- windows下python 安装igraph库报错 Cannot find the C core of igraph on this system using pkg-config.的解决
- scala学习之路:11.Scala常见操作一
- 那些年一起踩过的坑 — java 自动装箱拆箱问题
- NDK开发(1)——从无到有
- 关于express下session的几个注意事项
- 儿子和女儿——解释器和编译器的区别与联系
- 位运算,大小写转换
- mysql和php
- C 标准库——<cmath>/<math.h>
- yii2-user 扩展插件的一些坑
- 第一次上机实验-2
- 再谈EF Core内存数据库单元测试问题
- 输入输出流I/O
- python(5)-正则表达式
- 杭电oj 2027 统计元音
- 博客地址迁徙
- usb延长器比较
- 浅谈JS代码和OC代码的交互
- 树的应用 森林 并查集
- 方向是对的