[二分-最大化平均值]poj-2976

Dropping tests

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 15588		Accepted: 5440

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:

给出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(比率最后*100)

题解:

最大化平均值 : 01分数规划

最裸的01分数规划
设x[i]属于{0,1},表示第i个元组是否留下,p为比率,P为p的最大值,即比例的最大值(注意区分P和p)
则 p = sigma(ai*xi) / sigma(bi*ai),其中sigma(xi) = n-k; -> 表示有n-k个1
显然;对于所有可能取得的p的值,p <= P;
即对于所有可能的xi的组合,sigma(ai*xi) / sigma(bi*xi) <= P
即sigma(ai*xi) / sigma(bi*xi)的最大值就是等于P的
于是对于p sigma(ai*xi) - sigma(bi*xi*p) > 0;当 p < P;
当p = P时 ,sigma(ai*xi) - sigma(bi*xi*p) == 0;
当p > P,时 sigma(ai*xi) - sigma(bi*xi*p) < 0

我们要二分寻找p,使得p无限接近P. 于是当 max(sigma(ai*xi) - sigma(bi*xi*p)) >= 0 时说明此时满足情况,需要最大化向右搜
否则向左搜

代码:

/**
    最大化平均值 : 01分数规划
    给出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(比率最后*100)
    题解:最裸的01分数规划
    设x[i]属于{0,1},表示第i个元组是否留下,p为比率,P为p的最大值,即比例的最大值(注意区分P和p)
    则 p = sigma(ai*xi) / sigma(bi*ai),其中sigma(xi) = n-k; -> 表示有n-k个1
    显然;对于所有可能取得的p的值,p <= P;
    即对于所有可能的xi的组合,sigma(ai*xi) / sigma(bi*xi) <= P
    即sigma(ai*xi) / sigma(bi*xi)的最大值就是等于P的
    于是对于p sigma(ai*xi) - sigma(bi*xi*p) > 0;当 p < P;
    当p = P时 ,sigma(ai*xi) - sigma(bi*xi*p) == 0;
    当p > P,时 sigma(ai*xi) - sigma(bi*xi*p) < 0

    我们要二分寻找p,使得p无限接近P. 于是 当 max(sigma(ai*xi) - sigma(bi*xi*p)) >= 0 时 说明此时满足情况,需要最大化向右搜
    否则向左搜

*/
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn = 1e3+10;
const int inf = 1e9+10;
const double eps = 1e-6;
int arr[maxn],tot[maxn];
double ans[maxn];
int n,k;
bool check(double mid)
{
    for(int i=0;i<n;i++) {
        ans[i] = ((double)arr[i] - mid * tot[i]);
    }
    sort(ans,ans+n);
    double sum = 0;
    for(int i=k;i<n;i++) {
        sum += ans[i];
    }
    return sum >= 0;
}
int main()
{
    while(~scanf("%d%d",&n,&k),n+k)
    {
        for(int i=0;i<n;i++) scanf("%d",&arr[i]);
        for(int i=0;i<n;i++) scanf("%d",&tot[i]);
        double left = 0,right = inf;
        while(left + eps < right) {
            double mid = (left + right) / 2;
            if(check(mid)) left = mid;
            else right = mid;
        }
        right = right * 100;
        printf("%.0f\n",right);
    }
}

[二分-最大化平均值]poj-2976

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