牛客 A.gpa(01分数规划)

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

 

 

 

题目大意：

给定 n 门课以及它们的学分和绩点,定义总绩点是所有课的加权平均数,给定一个数 k, 你可以删除最多 k 门课,求你的总绩点最大能到多少
 

 

 

 

分析：

上面是牛客的官方题解，其实就是移项， 然后按照 c[i] - D 排一下序 然后求前几个的和

 

 

 

 

AC代码：

 /*************************************************
       Author       :	NIYOUDUOGAO
       Last modified:	2018-08-03 10:21
       Email        :	[e-mail]
       Filename     :	t1.cpp
 *************************************************/
#include <bits/stdc++.h>
#define mset(a, x) memset(a, x, sizeof(a))
using namespace std;
typedef long long ll;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7;
const int N = 1e7 + 5;
int s[N], c[N];
double t[N];
int n, k;
struct node {
	int s, c;
}a[N];
bool cmp(int x, int y) {
	return x > y;
} 
bool judge(double d) {
	double res = 0;
	for (int i = 0; i < n; i++) {
		t[i] = a[i].s * (a[i].c - d);
	}
	sort(t, t + n, cmp);
	for (int i = 0; i < n - k; i++) {
		res += t[i];
	}
	if (res >= 0) return 1;
	return 0;
}
int main() {
	scanf("%d%d", &n, &k); 
	for (int i = 0; i < n; i++) {
		scanf("%d", &a[i].s);
	}
	for (int i = 0; i < n; i++) {
		scanf("%d", &a[i].c);
	}
	double l = 0, r = 1001;
	double ans = 0;
	while (r - l >= 1e-8) {
		double mid = (l + r) / 2.0;
		if (judge(mid)) {
			ans = mid;
			l = mid + 0.000000001;
		} else {
			r = mid - 0.000000001;
		}
	}
	printf ("%.11lf\n", ans);
	return 0;
}