Poj 2976 Dropping tests【01分数规划+贪心】

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Dropping tests

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 11942 Accepted: 4169

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

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

题目大意：

给你N道题，每道题有两个元素：Ai，Bi，你必须去删除K道题，使得最终剩余的题的.值最大

思路：

1、很显然问题属于01分数规划问题。

当有求：Σ（a【i】）/Σ（b【i】）的最大（小）值的时候，我们可以将问题转化变成减法：

设定函数F（L）=Σ（a【i】）-L*Σ（b【i】）；

若此时F（L）>=0，那么肯定有Σ（a【i】）-L*Σ（b【i】）>=0，那么就有：Σ（a【i】）/Σ（b【i】）>=L;

根据化出的等式可以得知，我们有比L更大。那么我们这里可以二分枚举这个L，同时也就是在二分枚举最终的答案。

若有F（L）>=0，我们要增大答案，相反减小答案即可。

2、对于F（L）的计算，我们肯定希望F（L）尽可能的大，考虑每一个元素对最终结果会增大：Ai-L*Bi.那么我们对其贡献的结果从大到小排序即可。

然后贪心的取前n-k个元素，使得结果尽可能的大。

4、注意结果要四舍五入。

Ac代码：

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;struct node{    double x,y,L;}a[150000];int n,k;int cmp(node a,node b){    return a.x-a.L*a.y>b.x-b.L*b.y;}int Slove(double L){    double sum=0;    for(int i=0;i<n;i++)a[i].L=L;    sort(a,a+n,cmp);    for(int i=0;i<n-k;i++)    {        sum+=a[i].x-L*a[i].y;    }    if(sum>=0)return 1;    else return 0;}int main(){    while(~scanf("%d%d",&n,&k))    {        if(n==0&&k==0)break;        for(int i=0;i<n;i++)scanf("%lf",&a[i].x);        for(int i=0;i<n;i++)scanf("%lf",&a[i].y);        double ans=0;        double l=0;        double r=1000000000000000000;        for(int i=0;i<80;i++)        {            double mid=(l+r)/2;            if(Slove(mid)==1)            {                ans=mid*100;                l=mid;            }            else r=mid;        }        printf("%d\n",(int)(ans+0.5));    }}

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