链接：https://www.nowcoder.com/acm/contest/143/A
来源：牛客网
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
题意：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 1 <=n <=10^5
分析：假设最大总绩点为D，则题中式子可以转化成∑s[i](c[i]-D)=0
观察式子容易看出这是一个分数规划问题，我们可以通过二分D来解答，使左边式子结果最接近0的D即是最大总绩点D
以后类似的分数或者方程式不等式求解问题都可以用这种方式求解
01分数参考博客：https://blog.csdn.net/hhaile/article/details/8883652
AC代码：
#include <map>
#include <set>
#include <stack>
#include <cmath>
#include <queue>
#include <cstdio>
#include <vector>
#include <string>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <algorithm>
#define ls (r<<1)
#define rs (r<<1|1)
#define debug(a) cout << #a << " " << a << endl
using namespace std;
typedef long long ll;
const ll maxn = 1e5 + 10;
const ll mod = 1e9 + 7;
ll s[maxn], c[maxn], n, k;
double le, ri, t[maxn];
bool check( double x ) {
for( ll i = 1; i <= n; i ++ ) {
t[i] = s[i]*(c[i]-x);
}
sort( t+1, t+n+1 );
double tmp = 0.0;
for( ll i = k+1; i <= n; i ++ ) {
tmp += t[i];
}
return tmp>0;
}
int main() {
ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
cin >> n >> k;
for( ll i = 1; i <= n; i ++ ) {
cin >> s[i];
}
for( ll i = 1; i <= n; i ++ ) {
cin >> c[i];
ri = max( ri, (double)c[i] );
}
double mid;
while( ri-le > 1e-6 ) {
mid = (le+ri)/2;
//debug(mid);
if( check(mid) ) {
le = mid;
} else {
ri = mid;
}
}
printf("%.10lf
",mid);
return 0;
}