Dropping tests [二分]

题意：给定n个组，每个组包括a，b两个值，问除去k个不选，能得到的$100*\frac{\sum{a_i}}{\sum{b_i}}$的最大值是多少。
思路：二分这个东西$\frac{\sum{a_i}}{\sum{b_i}}$，然后判断即可
二分的单调性：这个很明显，如果能得到结果满足$>=x$，那么对于所有的$x_i<x$一定有值满足大于等于$x_i$(废话)
二分的check(mid)：要满足$\frac{\sum{a_i}}{\sum{b_i}}>=mid$则$\sum(a_i-mid*b_i)>=0$,因为有k个不用，对这个值排个序，得到的结果从大到小选，与0比较即可。

#include<iostream>
#include<string>
#include<cstdio>
#include<cstring>
#include<bitset>
#include<algorithm>
#include<map>
#include<set>
#include<queue>
#include<vector>
#include<cstdlib>
#include<list>
#include<stack>
#include<cmath>
#include<iomanip>
using namespace std;
//#pragma comment(linker, "/STACK:1024000000,1024000000")
typedef long long LL;
void debug() {cout << "ok running!" << endl;}
int n, k;
struct node
{
    int v, w, id;
    double x;
    bool operator < (node& b)
    {
        return x > b.x;
    }
} a[100005];
bool check(double mid)
{
    for(int i = 0; i < n; ++i)
        a[i].x = a[i].v - mid*a[i].w;
    sort(a, a+n);
    double sum = 0;
    for(int i = 0; i < n-k; ++i)
        sum += a[i].x;
    if(sum >= 0)
        return 1;
    else return 0;
}
int main()
{
    ios::sync_with_stdio(false);
    #ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
    #endif // ONLINE_JUDGE
    while(cin >> n >> k)
    {
        if(n == 0 && k == 0)
            break;
        for(int i = 0; i < n; ++i)
        {
            a[i].id = i+1;
            cin >> a[i].v;
        }
        for(int i = 0; i < n; ++i)
            cin >> a[i].w;

        double l = 0.0, r = 999999999.0;
        while(r - l > 1e-4)
        {
            double mid = (l+r) / 2;
            //cout << mid << endl;
            if(check(mid))
                l = mid;
            else r = mid;
        }
        cout << (int)((l+r)/2*100+0.5) << endl;
    }
    return 0;
}