NC17400 gpa

题目

题目描述

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.

输入描述

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]

输出描述

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

示例1

输入

3 1
1 2 3
3 2 1

输出

2.33333333333

说明

Delete the third course and the final score is $\frac{2*2+3*1}{2+1}=\frac{7}{3}$

备注

$1≤ n≤ 10^5$

$0≤ k < n$

$1≤ s[i],c[i] ≤ 10^3$

题解

知识点：01分数规划。

删去 $k$ 个课程，那剩下取 $n-k$ 个课程。答案的可行性函数符合单调性，且答案是唯一零点，于是二分答案，有如下公式：

$$\begin{aligned} \frac{\sum s[i]c[i]}{\sum s[i]} &\geq mid\\ \sum (s[i]c[i] - mid \cdot s[i]) &\geq 0 \end{aligned}$$

因此每次检验对 $s[i]c[i]-mid\cdot s[i]$ 从大到小排序，取前 $n-k$ 个加和，观察是否满足不等式，来决定舍弃区间。

注意精度开大一次方。

时间复杂度 $O(n \log n + k)$

空间复杂度 $O(n)$

代码

#include <bits/stdc++.h>
 
using namespace std;
 
double s[100007], c[100007], sc[100007];
int n, k;
 
bool check(double mid) {
    for (int i = 0;i < n;i++) sc[i] = s[i] * c[i] - mid * s[i];
    sort(sc, sc + n, [&](double a, double b) {return a > b;});
    double sum = 0;
    for (int i = 0;i < n - k;i++) sum += sc[i];
    return sum >= 0;
}
 
int main() {
    std::ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    cin >> n >> k;
    for (int i = 0;i < n;i++) cin >> s[i];
    for (int i = 0;i < n;i++) cin >> c[i];
    double l = 0, r = 1e3;
    while (r - l >= 1e-6) {
        double mid = (l + r) / 2;
        if (check(mid)) l = mid;
        else r = mid;
    }
    cout << fixed << setprecision(6) << l << '\n';
    return 0;
}