链接:https://www.nowcoder.com/acm/contest/143/A
来源:牛客网
时间限制:C/C++ 1秒,其他语言2秒
空间限制:C/C++ 262144K,其他语言524288K
Special Judge, 64bit IO Format: %lld
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
3 1 1 2 3 3 2 1
输出
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
题意:显而易见,这道题用到了01规划,通俗点讲就是有n种物品,每个物品都有a[i],b[i]两种不同的情况,选择n-k个元素要求
最大,反观这道题,直接将数据带入即可
我是使用的01规划中的二分法写的,但是这道题有一点要改进的就是至多删除k个元素,表示可以删除<=k个元素然后取最大值
这是我看的关于01规划算法的介绍https://blog.csdn.net/hzoi_ztx/article/details/54898323
主要公式:求max{f(r)}
#include <iostream>//二分法
#include <stdio.h>
#include <algorithm>
using namespace std;
int n,k,a[100005],b[100005];
double ps[100005];
const double eps=1e-5;
bool ok(double h)//f(r)=xi*a[i]*b[i]的和-xi*a[i]的和*r=(a[i]*b[i]的和-a[i]的和*r)*xi;为求最大值将xi全部为1,之后再将要删去的定为0.(xi={0,1})
{
for(int i=1;i<=n;i++) ps[i]=a[i]*b[i]-h*a[i];//ps数组表示横坐标为h是纵坐标的值后再将其全部加起来得到ans
sort(ps+1,ps+1+n);
double ans=0; //ans表示max{f(r)}即垂直于x轴坐标为h的直线与ps所代表的直线方程的交点;如果ans大于0则表示最大值在右边,否则在左边
for(int i=n;i>=k+1;i--) ans+=ps[i];
for(int i=2;i<=k;i++)
if(ps[i]>0)ans+=ps[i];
return ans>-eps;
}
void sol()
{
for(int i=1;i<=n;i++) scanf("%d",a+i);
for(int i=1;i<=n;i++) scanf("%d",b+i);
double l=0,r=10000;
while(r-l>eps)
{
double mid=(l+r)/2;
if(ok(mid)) l=mid; else r=mid;
}
printf("%.11lf\n",l);
}
int main()
{
scanf("%d%d",&n,&k);
sol();
}