首页 > 学院 > 开发设计 > 正文

POJ2976-Dropping tests-最大化平均值

2019-11-08 18:30:32
字体:
来源:转载
供稿:网友

原题链接 Dropping tests Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 11056 Accepted: 3857 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 0 Sample Output

83 100 Hint

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

Stanford Local 2005 题意:从n门成绩中取出k门成绩,剩下的成绩算加权平均数(如题),求最大的平均数(四舍五入) 思路:二分平均数,这k门成绩单位均值的计算方式与x满足这样的关系这里写图片描述 通过变形可以得到 这里写图片描述 所以我们只需要按照这个式子进行处理然后排序找到后n-k位加和看是不是大于等于0即可知道该平均数是否符合条件

#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>using namespace std;const double INF = 0x3f3f3f3f;const double EPS = 1.0e-4;//10^-4const int maxn = 1010;int a[maxn],b[maxn],n,c;double t[maxn];bool _can(double x){ for(int i=0;i<n;i++) t[i] = a[i] - x*b[i]; sort(t,t+n); double sum = 0; for(int i=c;i<n;i++){ sum += t[i]; } return sum>=0.0;}int main(){ while(scanf("%d%d",&n,&c)==2 && n+c!=0){//从n个中取出c个,剩下的求加权平均数 for(int i=0;i<n;i++) scanf("%d",&a[i]); for(int i=0;i<n;i++) scanf("%d",&b[i]); double l=0,r=1.0; while(r-l>EPS){//这里EPS=1.0e-3的话就会是错的,但是1.0e-4就是对的 double mid = (l+r)/2; if(_can(mid)) l=mid; else r=mid; } printf("%d/n",(int)(l*100+0.5)); } return 0;}
发表评论 共有条评论
用户名: 密码:
验证码: 匿名发表