01分数规划POJ2976（简单模板题）

Dropping tests

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 7276		Accepted: 2523

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_ifor 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 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).

题意：给定n个二元组(a,b)，删除k个二元组，使得剩下的a元素之和与b元素之和的比率最大，最后的比率乘于100，然后输出跟最大比率最接近的整数

分析：设r=sigma(ai*xi)/sigma(bi*xi);其中xi={0,1},sigma(xi)=n-k,设R为最优值，

即：r<=R

即：sigma(ai*xi)/sigma(bi*xi)<=R

即：sigma(ai*xi)-sigma(R*bi*xi)<=0

也就是说sigma(ai*xi)-sigma(R*bi*xi)的最大值为0，

等价于 sigma((ai-R*bi)*xi)的最大值等于0；

因为h(r)=sigma((ai-R*bi)*xi)为单调递减函数，

所以可以二分求

 

#include"stdio.h"

#include"algorithm"

#include"string.h"

#include"iostream"

#include"queue"

#include"map"

#include"stack"

#include"cmath"

#include"vector"

#include"string"

#define M 1009

#define N 20003

#define eps 1e-7

#define mod 123456

#define inf 100000000

using namespace std;

int a[M],b[M],n,k;

double s[M];

int cmp(double a,double b)

{

    return a>b;

}

double fun(int n,double r)

{

    for(int i=1;i<=n;i++)

        s[i]=a[i]-r*b[i];

    sort(s+1,s+n+1,cmp);

    double sum=0;

    for(int i=1;i<=n-k;i++)

        sum+=s[i];

    return sum;

}

int main()

{

    while(scanf("%d%d",&n,&k),n||k)

    {

        double l=0,r=0,mid=0;

        for(int i=1;i<=n;i++)

        {

            scanf("%d",&a[i]);

            r+=a[i];

        }

        for(int i=1;i<=n;i++)

            scanf("%d",&b[i]);

        while(r-l>eps)

        {

            mid=(l+r)/2;

            double msg=fun(n,mid);

            if(msg<0)

            {

                r=mid;

            }

            else

            {

                l=mid;

            }

        }

        printf("%.0lf\n",r*100);

    }

    return 0;

}