Description
有nnn门课程,第iii门学分为sis_isi,得分为cic_ici,所选课程的总gpagpagpa为∑sici∑si\frac{\sum s_ic_i}{\sum s_i}∑si∑sici,问删除kkk门课程后剩余课程gpagpagpa的最大值
Input
第一行两个整数n,kn,kn,k,之后输入nnn个整数sis_isi,最后输入nnn个整数cic_ici
(1≤n≤105,0≤k<n,1≤si,ci≤103)(1\le n\le 10^5,0\le k<n,1\le s_i,c_i\le 10^3)(1≤n≤105,0≤k<n,1≤si,ci≤103)
Output
输出最大gpagpagpa
Sample Input
3 1
1 2 3
3 2 1
Sample Output
2.33333333333
Solution
010101分数规划,二分答案xxx,那么问题转化为删去kkk个元素使得∑(sici−xsi)≥0\sum (s_ic_i-xs_i)\ge 0∑(sici−xsi)≥0,排完序选最小的kkk个删掉即可
Code
#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;
#define maxn 100005
#define eps 1e-8
int a[maxn],b[maxn];
double d[maxn];
int n,k;
int sgn(double x)
{
if(fabs(x)<eps)return 0;
if(x>eps)return 1;
return -1;
}
double C(double x)
{
double ans=0;
for(int i=0;i<n;i++)d[i]=1.0*a[i]-x*b[i],ans+=d[i];
sort(d,d+n);
for(int i=0;i<k;i++)ans-=d[i];
return sgn(ans)>=0;
}
int main()
{
while(~scanf("%d%d",&n,&k))
{
for(int i=0;i<n;i++)scanf("%d",&b[i]);
for(int i=0;i<n;i++)scanf("%d",&a[i]),a[i]*=b[i];
double l=0,r=1000,ans=-1;
int T=30;
while(T--)
{
double mid=0.5*(l+r);
if(C(mid))l=ans=mid;
else r=mid;
}
printf("%.6f\n",ans);
}
return 0;
}