POJ-2976：Dropping tests【01分数规划】_pojdropping tests time limit: 1000ms-优快云博客

本文介绍了一个算法问题，即如何通过放弃部分考试成绩来最大化累计平均分。通过使用01分数规划的方法解决了这一问题，并详细解释了算法实现过程及注意事项。

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Dropping tests

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 12059		Accepted: 4212

Description

In a certain course, you take n tests. If you get a_i out of b_i 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 a_i for all i. The third line contains n positive integers indicating b_i for all i. It is guaranteed that 0 ≤ a_i ≤ b_i ≤ 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

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).

题意：从n对（a，b）中选择其中的n-k对，使（Σa[i]/Σb[i]）*100的值最大。

分析：01分数规划，注意精度问题，如果写成ans=l=mid，输出printf("%.0f\n",ans*100);是错的……wrong了很多次，才找到这里有问题。要写成l=mid，输出是l*100就可以了。

#include<stdio.h>
#include<algorithm>
#define eps 1e-9
using namespace std;
const int maxn=1111;
int n,k;
double a[maxn],b[maxn];
double x[maxn];
int cmp(double a,double b)
{
    return a>b;
}
int check(double l)
{
    for(int i=0;i<n;i++)
        x[i]=a[i]-l*b[i];
    sort(x,x+n,cmp);
    double sum=0;
    for(int i=0;i<n-k;i++)
        sum+=x[i];
    return sum>=0;
}
int main()
{
    while(scanf("%d%d",&n,&k),n+k)
    {
        for(int i=0;i<n;i++)
            scanf("%lf",&a[i]);
        for(int i=0;i<n;i++)
            scanf("%lf",&b[i]);
        double l=0,r=1.0;
        while(r-l>=eps)
        {
            double mid=(l+r)/2.0;
            if(check(mid))
                l=mid;
            else
                r=mid;
        }
        printf("%.0f\n",l*100);
    }
    return 0;
}