Dropping tests POJ - 2976 最大化平均值。

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 npositive 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).

有N次考试，每次考试有 b[i]个题目，但是知道每次能作出的题目 a[i], 然后要求去掉 k 个，求得到的 a/b*100 最大；

这个就是最大化平均值，就是挑战书上的P143，选出K个物品使得单位重量的价值最大。都是一个思路。

先枚举 x ，，就是结果，

然后根据 a-x*b 从大到小排序，选出要拿的物品。结果大于等于0就可行。

具体证明看书上。

#include<bits/stdc++.h>
using namespace std;
const int MAXN = 1009;
struct node{
    int a,b;
}p[MAXN];
int n,k;
double y[MAXN];

bool judge(double x){
    x = x/100;
    for (int i = 0; i < n; i++)
        y[i] = p[i].a - x*p[i].b;
    sort(y,y+n);
    double tot = 0;
    for (int i = n-1; i > k-1; i--){
        tot += y[i];
    }
    if (tot >= 0) return 1; else return 0;
}

int main() {
    while(scanf("%d%d",&n,&k)==2) {
        if (n == 0) break;

        for (int i = 0; i < n; i++)
            scanf("%d", &p[i].a);
        for (int i = 0; i < n; i++)
            scanf("%d", &p[i].b);
        double l = 0, r = 100,mid;
        while(r-l > 0.0001){
            mid = (l + r)/2;
            if (judge(mid)) l = mid; else r = mid;
        }
        int ans;
       // printf("%.5lf  ",l);
        ans = (l+0.5);
        printf("%d\n",ans);
    }
    return 0;
}

Dropping tests POJ - 2976 最大化平均值。

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