[ 2018牛客网暑期ACM多校训练营（第五场）A题] gpa

Problem Description
Kanade selected n courses in the university. The academic credit of the i-th course is s[i] and the
score of the i-th course is c[i].
At the university where she attended, the final score of her is $\frac{\sum s[i]c[i] }{\sum s[i]}$
Now she can delete at most k courses and she want to know what the highest final score that can
get.

Input
The first line has two positive integers n,k
The second line has n positive integers s[i]
The third line has n positive integers c[i]

Output
Output the highest final score, your answer is correct if and only if the absolute error with the
standard answer is no more than $10^{-5}$.

Note
1≤ n≤ 105
0≤ k < n
1≤ s[i],c[i] ≤ 103

Sample Input
3 1
1 2 3
3 2 1
Sample Output
2.33333333333
Hint
Delete the third course and the final score is $\frac{2 * 2 + 3 * 1}{2 + 1} = \frac{7}{3}$.

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题目大意：有n个学生，给出每个学生的学分和绩点，求从中去掉k个学生后剩下学生的“总绩点”的最大值。

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分析： 01分数规划裸题。设“总绩点”最大值为x，$\frac{\sum s[i] * c[i]}{\sum s[i]} = x$ → $\sum (s[i] * c[i] - x * c[i]) = 0$。因为[0-x)间一定有$\sum (s[i] * c[i] - x * c[i]) > 0$，(x, max_right]间一定有$\sum (s[i] * c[i] - x * c[i]) < 0$。故可二分x得到答案。复杂度为O($nlog_{2}n * log_{2}(1000 * 1e6)$)。

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代码：

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 1e5;
int n, k;
struct fct{int n, d;} s[N];
double val;
bool cmp(fct a, fct b)
{
    return a.n - a.d * val > b.n - b.d * val;
}
int main()
{
    cin >> n >> k;
    k = n - k;
    for (int i = 0; i < n; i++)
        cin >> s[i].d;
    for (int i = 0; i < n; i++)
    {
        cin >> s[i].n;
        s[i].n *= s[i].d;
    }
    double l = 0, r = 1000;
    while (r - l > 1e-6)
    {
        val = (l + r) / 2;
        sort(s, s + n, cmp);
        double ans = 0;
        for (int i = 0; i < k; i++)
            ans += s[i].n - val * s[i].d;
        if (ans >= 0)
            l = val;
        else
            r = val;
    }
    cout << setprecision(5) << fixed << val << endl;
    return 0;
}