POJ 2976 Dropping tests（01分数规划）

Dropping tests

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 15067		Accepted: 5263

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 1
5 0 2
5 1 6
4 2
1 2 7 9
5 6 7 9
0 0

Sample Output

83
100

题意

给出n个二元组$(a_1,b_1)(a_2,b_2)...(a_n,b_n)$，从中选出$(n-k+1)$个二元组，使得选出的这k个二元组$c=frac{a_1+a_2+...+a_k}{b_1+b_2+...+b_k}$最大。

分析

01分数规划。

对于原问题，并不好分析，将问题转化一下，转化成判定性问题。

原题中使得c最大，假设c就是答案，那么有：

$a_1+a_2+...+a_k=c*(b_1+b_2+...+b_k)$
$a_1+a_2+...+a_k=c*b_1+c*b_2+...+c*b_k$
$(a_1-c*b_1)+(a_2-c*b_2)...+(a_k-c*b_k)=0$

所以我们可以二分一个c，判断是否可行。

设新数组$d[i]=a[i]-c*b[i]$，从d数组中挑出$(n-k+1)$最大的数，如果大于等于0，那么c满足，否则不满足。

复杂度$O(nlogn)$

code

#include<cstdio>
#include<algorithm>

using namespace std;

const double eps = 1e-8;
double a[1010],b[1010],c[1010];
int n,k;

bool check(double x) {
    for (int i=1; i<=n; ++i) 
        c[i] = a[i] - x * b[i];
    sort(c+1,c+n+1);
    double ans = 0.0;
    for (int i=k+1; i<=n; ++i) ans += c[i];
    return ans >= 0.0;
}
int main() {
    while (~scanf("%d%d",&n,&k)) {
        if (n==0 && k==0) break;
        for (int i=1; i<=n; ++i) scanf("%lf",&a[i]);
        for (int i=1; i<=n; ++i) scanf("%lf",&b[i]);
        double L = 0.0,R = 1.0,mid;
        while (R-L>eps) {
            mid = (L + R) / 2.0;
            if (check(mid)) L = mid;
            else R = mid;
        }
        printf("%.0lf
",L*100);
    }
    return 0;
}