标题：A、gpa | 时间限制：1 秒 | 内存限制：256M
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
备注: 1≤n≤10^5
0≤ k<n
1≤s[i],c[i] ≤10^3
示例 1
输入
3 1
1 2 3
3 2 1
输出
2.33333333333
题目大意：给出每门的课的学分与绩点，算总绩点，输出误差在10-5。
思路：比赛时死活过不了，全是wa，心态爆炸...我的方法确实是错的，不然应该是超时而不是wa。之前的想法是把学分除以绩点再排序，再找去除1-k课的最大情况。然而是错的，联想以前看过的文章，，应该是死在这儿了，，，
正确的做法：
所以二分找最大绩点就行了........(出题人说需要注意二分的上界不然会超时)心好累好累好累。
代码如下：
#include<iostream>
#include<algorithm>
#include<cstdio>
using namespace std;
long long n,k,i,j;
long long s[100050],c[100050];
double p[100050];
bool fun(double x)
{
for(i=0;i<n;i++)
{
p[i]=s[i]*c[i]-x*s[i];
}
sort(p,p+n);
double sum=0;
for(i=k;i<n;i++)
{
sum+=p[i];
}
return sum>0;
}
int main()
{
cin>>n>>k;
for(i=0;i<n;i++) cin>>s[i];
for(i=0;i<n;i++) cin>>c[i];
double l=0,r=5000000,mid;
for(j=0;j<80;j++)
{
mid=(l+r)/2;//二分找最大学分
if(fun(mid)) l=mid;
else r=mid;
}
printf("%.11lf\n",r);
return 0;
}