牛客网暑期ACM多校训练营（第五场）A gpa（01分数规划）

链接：https://www.nowcoder.com/acm/contest/143/A
来源：牛客网
 

时间限制：C/C++ 1秒，其他语言2秒
空间限制：C/C++ 262144K，其他语言524288K
Special Judge, 64bit IO Format: %lld

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 

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]

时间限制：C/C++ 1秒，其他语言2秒
空间限制：C/C++ 262144K，其他语言524288K
Special Judge, 64bit IO Format: %lld

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 

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

复制

3 1
1 2 3
3 2 1

输出

复制

2.33333333333

说明

Delete the third course and the final score is 

备注:

1≤ n≤ 105

0≤ k < n

1≤ s[i],c[i] ≤ 103

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

复制

3 1
1 2 3
3 2 1

输出

复制

2.33333333333

说明

Delete the third course and the final score is 

备注:

1≤ n≤ 105

0≤ k < n

1≤ s[i],c[i] ≤ 103

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define LL long long
#define clr(a) memset(a,0,sizeof(a))

#define eps 1e-12
const int MAXN = 1e5+10;
const int INF = 0x3f3f3f3f;
const int N = 1010;

int n,k;
double a[MAXN],b[MAXN],w[MAXN];

int main(){
    while(scanf("%d%d",&n,&k)!=EOF&&(n||k)){
        clr(a);clr(b);clr(w);
        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 = n,mid;
        double maxx = 0.0;
        while(r - l > eps){
            mid = (l + r) / 2;
            for(int i=1;i<=n;i++)
                w[i] = a[i] * b[i] - mid * a[i];
            sort(w+1,w+n+1);
            maxx = 0;
            for(int i=n;i>k;i--)
                maxx += w[i];
            if(maxx > eps)  l = mid;
            else r = mid;
        }
        printf("%lf\n",r);
    }
    return 0;
}