[二分答案]gpa

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

思路：不能将成绩最差的k门课去掉，因为一门课拖后腿的程度不仅与它的分数有关，也与它的学分多少有关，比如两门课，一门60，一门59，不能就说59的那门课拖后腿更多，如果59的那门课只占一个学分，而60的那门课占5个学分，那就不一样了；
但要肯定的是，尽可能去掉更多的课，所以要使结果最优要去掉k门课；可以考虑二分枚举答案(gpa)，计算出每门课对应的s*c-s*gpa,取该结果最高的n-k门课并求和，若>=0，则该gpa可以达到；
AC代码：

#include <iostream>
#include<cstdio>
#include<algorithm>
typedef long long ll;
using namespace std;

struct Course{
  ll s,c;
}course[100010];
double b[100010];
ll n,k;

bool ok(double a){
  for(ll i=1;i<=n;i++) b[i]=course[i].s*course[i].c*1.0-course[i].s*a*1.0;
  sort(b+1,b+1+n);
  double sum=0;
  for(ll i=n;i>=k+1;i--) sum+=b[i];
  return sum>=0;
}

int main()
{
    scanf("%lld%lld",&n,&k);
    for(ll i=1;i<=n;i++) scanf("%lld",&course[i].s);
    for(ll i=1;i<=n;i++) scanf("%lld",&course[i].c);
    double L=0,R=1000.0;
    while(R-L>1e-6){
        double mid=(L+R)*1.0/2.0;
        if(ok(mid)) L=mid;
        else R=mid;
    }
    printf("%.6f\n",R);
    return 0;
}