時間限制：C/C++ 1秒，其他語言2秒
空間限制：C/C++ 262144K，其他語言524288K
Special Judge, 64bit IO Format: %lld

題目描述

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 

Now she can delete at most k courses and she want to know what the highest final score that can get.

輸入描述:

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]

輸出描述:

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

輸入

複製

3 1
1 2 3
3 2 1

輸出

複製

2.33333333333

說明

Delete the third course and the final score is 

備註:

1≤ n≤ 105

0≤ k < n

1≤ s[i],c[i] ≤ 103

看著就是一臉“我是01分數規劃”的樣子

讓a[i]=s[i]c[i] b[i]=s[i]，就變成01分數規劃了

找前k大的可以用nth_element找

#include<bits/stdc++.h>
#define mp make_pair
#define fir first
#define se second
#define ll long long
#define pb push_back
using namespace std;
const int maxn=2e5+10;
const ll mod=1e9+7;
const int maxm=1e6+10;
const double eps=1e-7;
const int inf=0x3f3f3f3f;
const double pi = acos (-1.0);
int n,k;
pair<ll ,ll >p[maxn];
pair<ll,ll> p1[maxn];
double temp[maxn];
int check(double mid){
    for (int i=0;i<n;i++){
        temp[i]=p1[i+1].fir-mid*p1[i+1].se;
    }
    nth_element(temp,temp+n-k,temp+n);
    reverse(temp,temp+n);
    double sum=0;
    for (int i=0;i<k;i++)
        sum+=temp[i];
    return sum>0;
}
int main(){
    scanf("%d %d",&n,&k);
    k=n-k;
    for (int i=1;i<=n;i++){
        scanf("%lld",&p[i].fir);
    }
    for (int i=1;i<=n;i++){
        scanf("%lld",&p[i].se);
    }
    for (int i=1;i<=n;i++){
        p1[i].fir=p[i].fir*p[i].se;
        p1[i].se=p[i].fir;
    }
    double l=0,r=1000;
    for (int i=0;i<100;i++){
        double mid=(l+r)/2;
        if (check(mid)){
            l=mid;
        }
        else r=mid;
    }
    printf("%.14f\n",max(l,r));
    return 0;
}