POJ-2976：Dropping tests【01分数规划】

Dropping tests

Time Limit: 1000MS Memory Limit: 65536KTotal 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

3 15 0 25 1 64 21 2 7 95 6 7 90 0

Sample Output

83100

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-9using 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;}