牛客网暑期ACM多校训练营（第五场） gpa

版权声明：本文为博主原创文章，未经博主允许不得转载。 https://blog.csdn.net/gtuif/article/details/81814585

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

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


题意：删除k节课使gpa最高;

思路：二分答案，看sum(s[i]*(c[i] - ans)) >=0 （k+1<=i<=n）是否成立;(贪心是错的）

下面附上我的代码：

#pragma GCC optimize ("O3")
#include<cstdio>
#include<cmath>
#include<cstring>
#include<map>
#include<set>
#include<algorithm>
#include<iostream>
#include<vector>
#include<queue>
#include<stack>
#include<string>
#define LL long long
#define ull unsigned long long
#define INF 0x3f3f3f3f
#define PI acos(-1.0)
#define lson o<<1
#define rson o<<1|1
#define eps 1e-8
using namespace std;
const int maxn = 1e5 + 10;
double s[maxn];
double c[maxn];
double d[maxn];
int n, k;
double Min(double x, double y)
{
	if(x > y)
		return y;
	return x;
}
double Max(double x, double y)
{
	if(x > y)
		return x;
	return y;
}
int judge(double x)
{
	for(int i = 0; i < n; i++)
		d[i] = (c[i] - x) * s[i];
	sort(d, d + n);
	double sum = 0;
	for(int i = k; i < n; i++)
		sum += d[i];
	if(sum >= 0)
		return 1;
	return 0;
}
int main()
{
	cin >> n >> k;
	double Mi = -1, Ma = 1005; 
	for(int i = 0; i < n; i++)
		scanf("%lf", &s[i]);
	for(int i = 0; i < n; i++)
	{
		scanf("%lf", &c[i]);
		Mi = Min(Mi, c[i]);
		Ma = Max(Ma, c[i]);
	}
	double mid;
	while(Ma - Mi >= eps)
	{
		mid = (Ma + Mi) / 2;
		if(judge(mid))
			Mi = mid;
		else
			Ma = mid;
	}
	printf("%.7lf\n", Ma);
    return 0;
}