牛客暑假多校第五场A.gpa

一、题意

　　给出你的N门课程的考试成绩和所占的机电数目。允许你放弃K门课的成绩，要求你的平均学分绩最高能达到多少。

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.

二、题解

　　对于这道题直观的思路就是“猜一个目标学分绩点”之后验证在抛弃k门课的基础上，是否可以达到目标学分绩点。

　　对于每个目标学分绩点，对所有成绩进行一轮排序使得：拉低给定平均学分绩点最多的科目排在最前面，拉高给定平均学分绩点的成绩排在最后面。

　　则，该思路可以规避一些比较难想的复杂操作。

#include<bits/stdc++.h>
using namespace std;

#define ll long long
#define veci vector<int>

const int MAXN=;
const double EXP = 1e-;

double sum_credit;
double sum_mul;
double target_gread;

double arr[MAXN];
double brr[MAXN];
int courses[MAXN];
int n,k;

bool cmp_tar(int a,int b)
{
    double tmpa = (target_gread *(sum_credit - arr[a]) + brr[a]*arr[a])/sum_credit - target_gread;
    double tmpb = (target_gread * (sum_credit - arr[b]) + brr[b]*arr[b])/sum_credit - target_gread;
    return tmpa < tmpb;
}

bool check(double target)
{
    target_gread = target;
    sort(courses,courses+n,cmp_tar);

        // cout<<"seq1: ";
        // for(int i=0;i<n;++i)
        //     cout<<arr[courses[i]]<<" ";
        // cout<<endl<<"seq2: ";
        // for(int i=0;i<n;++i)
        //     cout<<brr[courses[i]]<<" ";
        // cout<<endl;

    double ans = ;
    double summ = ;
    for(int i=min(k,n-);i<n;++i)
    {
            // cout<<"arr: "<<arr[courses[i]]<<" brr: "<<brr[courses[i]]<<endl;
        ans += arr[courses[i]] * brr[courses[i]];
        summ += arr[courses[i]];
    }
// cout<<"check: "<<target<<" "<<ans<<" "<<summ<<endl;

    ans /= summ;
    return ans>=target;
}

double bin_search(double a,double b)
{

        // cout<<a<<" : "<<b<<endl;
    if(abs(a-b)<=EXP)
        return a;

    double mid = (a+b)/;
    if(check(mid))return bin_search(mid,b);
    else return bin_search(a,mid);
}

void init()
{
    sum_credit = sum_mul = ;
    for(int i=;i<n;++i)
        scanf("%lf",&arr[i]);
    for(int i=;i<n;++i)
    {
        sum_credit += arr[i];
        scanf("%lf",&brr[i]);
        sum_mul += arr[i]*brr[i];
        courses[i] = i;
    }
    printf("%.6lf\n",bin_search(,));
}

int main(){

    cin>>n>>k;
    init();

    return ;
}