[poj P2976] Dropping tests
[poj P2976] Dropping tests
Time Limit: 1000MS Memory Limit: 65536K
Description
In a certain course, you take n tests. If you get ai out of bi questions correct on test i, your cumulative average is defined to be
.
Given your test scores and a positive integer k, determine how high you can make your cumulative average if you are allowed to drop any k of your test scores.
Suppose you take 3 tests with scores of 5/5, 0/1, and 2/6. Without dropping any tests, your cumulative average is
. However, if you drop the third test, your cumulative average becomes
.
Input
The input test file will contain multiple test cases, each containing exactly three lines. The first line contains two integers, 1 ≤ n ≤ 1000 and 0 ≤ k < n. The second line contains n integers indicating ai for all i. The third line contains n positive integers indicating bi for all i. It is guaranteed that 0 ≤ ai ≤ bi ≤ 1, 000, 000, 000. The end-of-file is marked by a test case with n = k = 0 and should not be processed.
Output
For each test case, write a single line with the highest cumulative average possible after dropping k of the given test scores. The average should be rounded to the nearest integer.
Sample Input
3 1 5 0 2 5 1 6 4 2 1 2 7 9 5 6 7 9 0 0Sample Output
83 100Hint
To avoid ambiguities due to rounding errors, the judge tests have been constructed so that all answers are at least 0.001 away from a decision boundary (i.e., you can assume that the average is never 83.4997).
Source
01分数规划入门题。
这个不等式会经常看到:sigma(ai)/sigma(bi)>=(或<=)k
对于这题来说就是,找出最大的k,使得100*sigma(ai)/sigma(bi)>=k。
我们尝试化简上式。
因为bi>=0,所以sigma(bi)>=0,100*sigma(ai)>=k*sigma(bi)
100*sigma(ai)-k*sigma(bi)>=0
sigma(100*ai-k*bi)>=0
那么,我们只要先枚举一个k,将100*ai-k*bi作为关键字排序,再选出前n-k大的,判断一下sum是否非负就行了。
然后我们发现,上式满足单调性,枚举可以改为用二分。所以总复杂度是O(nlog分数)。
然后这一题我先开大100倍,然后最后再缩小100倍,但是这也会有精度误差。
总之,开大的倍数越大,误差越小,但效率越低,但也低不到哪里去qwq
code:
#include<cstdio> #include<cstring> #include<algorithm> #define LL long long #define M ((L)+(R)>>1) using namespace std; ,lim=; int n,k,L,R,ans,final; LL sum; struct ob {LL a,b,c;}o[N]; bool cmp(ob x,ob y) {return x.c>y.c;} bool jug(LL lv) { ; i<=n; i++) o[i].c=o[i].a*lim-lv*o[i].b; sort(o+,o++n,cmp); sum=; ; i<=n-k; i++) sum+=o[i].c; ; } int main() { while (scanf("%d%d",&n,&k)!=EOF,n|k) { ; i<=n; i++) scanf("%lld",&o[i].a); ; i<=n; i++) scanf("%lld",&o[i].b); ,R=lim; L<=R; ) ; ; final=ans/; ans%=; ) final+=; printf("%d\n",final); } ; }
[poj P2976] Dropping tests的更多相关文章
- POJ - 2976 Dropping tests && 0/1 分数规划
POJ - 2976 Dropping tests 你有 \(n\) 次考试成绩, 定义考试平均成绩为 \[\frac{\sum_{i = 1}^{n} a_{i}}{\sum_{i = 1}^{n} ...
- 二分算法的应用——最大化平均值 POJ 2976 Dropping tests
最大化平均值 有n个物品的重量和价值分别wi 和 vi.从中选出 k 个物品使得 单位重量 的价值最大. 限制条件: <= k <= n <= ^ <= w_i <= v ...
- POJ 2976 Dropping tests(01分数规划)
Dropping tests Time Limit: 1000MS Memory Limit: 65536K Total Submissions:17069 Accepted: 5925 De ...
- POJ 2976 Dropping tests 【01分数规划+二分】
题目链接:http://poj.org/problem?id=2976 Dropping tests Time Limit: 1000MS Memory Limit: 65536K Total S ...
- POJ 2976 Dropping tests(01分数规划入门)
Dropping tests Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 11367 Accepted: 3962 D ...
- POJ 2976 Dropping tests 01分数规划 模板
Dropping tests Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 6373 Accepted: 2198 ...
- POJ 2976 Dropping tests (0/1分数规划)
Dropping tests Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 4654 Accepted: 1587 De ...
- Poj 2976 Dropping tests(01分数规划 牛顿迭代)
Dropping tests Time Limit: 1000MS Memory Limit: 65536K Description In a certain course, you take n t ...
- poj 2976 Dropping tests 二分搜索+精度处理
Dropping tests Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 8349 Accepted: 2919 De ...
随机推荐
- keras框架 反复调用model 模型 出错
Cannot interpret feed_dict key as Tensor: Tensor Tensor("Placeholder_8:0", shape=(3, 3, 12 ...
- laravel----------php7.0.12 laravel 链接sqlserver数据库
https://www.microsoft.com/en-us/download/details.aspx?id=20098 下载最后一个,然后这个工具可以将dll扩展下载下来,选择一个空白的文件夹就 ...
- Vue学习1:实例及生命周期
<!DOCTYPE html> <html lang="en"> <head> <meta charset="UTF-8&quo ...
- C#中的装箱(inboxing)和拆箱(unboxing)(简单理解)
装箱和拆箱是值类型和引用类型之间相互转换是要执行的操作. 装箱:将一个值类型隐式地转换成一个object类型,或把这个值类型转换成一个被该值类型应用的接口类型,把一个值类型的值装箱,就是创建一个ob ...
- centos7.5 修改网卡名称
1.修改网卡配置文件中名称信息 vim /etc/sysconfig/network-scripts/ifcfg-ens33 将其中的名称为ens33的改为eth0 ,并将uuid删除以便后面克隆 2 ...
- 【awk】提取文件第一列
生信数据文件一般是按列分开的,如果我们只想简单的提取一列而不是费尽周折写个程序提取哪一列的话,awk作为一个非常好用的文档处理工具,我们现在来简单看一下他的一些功能: awk '{print $1}' ...
- Dart基础-泛型和库
https://blog.csdn.net/hekaiyou/article/details/46774727
- mac忘记操作密码
转载于:https://blog.csdn.net/wu110112/article/details/70312987 如果忘记mac登陆密码应该如何处理呢? 这里大家请勿着急,我来帮大家解决这个问题 ...
- 微信小程序,天气预报(百度地图开放平台API)
小程序看似一种全新的东西,但好在基本上就是曾经HTML,CSS,JS的一个微变版本. 语法和之前一样.只是一些用法和名字(标签)发生了一些变化. 小程序主要就四种扩展名的文件:js,json,wxml ...
- vue style background
vue 动态加载背景图 :style="{backgroundImage: 'url('+ item.imgList[0] +')',backgroundRepeat:'no-repeat' ...