A gpa（0-1分数规划）

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

0-1分数规划
r=(t1*x1+t2*x2+…+tn*xn)/(c1*x1+c2*x2+…+cn*xn)
给定t[1-n],c[1-n]，求x[1-n]使得sigma（xi）=k且r最大（小）
为了让r最大，设z(r)= (t1*x1+ .. +tn*xn)-r * (c1*x1+..+cn*xn)
假设r的最优值为R，则：
z(r)<0,当且仅当r>R;
z(r)=0,当且仅当r=R;
z(r)>0,当且仅当r

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#define inf 0x3f3f3f3f
#define LL long long

using namespace std;
const int maxn=100005;
int n,k;
int s[maxn],c[maxn];
double f[maxn];

int check(double m)
{
    for(int i=0;idouble ans=0;
    for(int i=k;ireturn ans>=0;
}

int main()
{
    while(scanf("%d%d",&n,&k)==2){
        for(int i=0;iscanf("%d",&s[i]);
        }
        for(int i=0;iscanf("%d",&c[i]);
        }
        for(int i=0;iint t=s[i];
            s[i]=s[i]*c[i];
            c[i]=t;
        }
        double l=0,r=inf;
        while(r-l>0.000001){
            double mid=(l+r)/2;
            if(check(mid)){
                l=mid;
            }
            else{
                r=mid;
            }
        }
        printf("%.11f\n",l);
    }
    return 0;
}