[POJ2976]Dropping tests（01分数规划，二分）

题目链接：http://poj.org/problem?id=2976

题意：你有n个成绩要从中删除k个，使得100*(∑ai/∑bi)值最大。

这是个简单的01分数规划题，关于01分数规划：http://www.cnblogs.com/perseawe/archive/2012/05/03/01fsgh.html。

通过数学分析获得一个目标式的变形后，可以知道F(L)=∑ai*xi-L*∑bi*xi，使F(L)=0就是使L最大化。我们枚举L后获得ai*xi-L*bi*xi的值，对这个值从大到小排序，再去算前n-k个的∑ai*xi和∑bi*xi，看看∑ai*xi/∑bi*xi是否大于二分得到的L，如果大于则更新左值，否则更新右值。

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*/
#include <algorithm>
#include <iostream>
#include <iomanip>
#include <cstring>
#include <climits>
#include <complex>
#include <fstream>
#include <cassert>
#include <cstdio>
#include <bitset>
#include <vector>
#include <deque>
#include <queue>
#include <stack>
#include <ctime>
#include <set>
#include <map>
#include <cmath>
using namespace std;
#define fr first
#define sc second
#define cl clear
#define BUG puts("here!!!")
#define W(a) while(a--)
#define pb(a) push_back(a)
#define Rint(a) scanf("%d", &a)
#define Rs(a) scanf("%s", a)
#define Cin(a) cin >> a
#define FRead() freopen("in", "r", stdin)
#define FWrite() freopen("out", "w", stdout)
#define Rep(i, len) for(int i = 0; i < (len); i++)
#define For(i, a, len) for(int i = (a); i < (len); i++)
#define Cls(a) memset((a), 0, sizeof(a))
#define Clr(a, x) memset((a), (x), sizeof(a))
#define Full(a) memset((a), 0x7f7f7f, sizeof(a))
#define lrt rt << 1
#define rrt rt << 1 | 1
#define pi 3.14159265359
#define RT return
#define lowbit(x) x & (-x)
#define onecnt(x) __builtin_popcount(x)
typedef long long LL;
typedef long double LD;
typedef unsigned long long ULL;
typedef pair<int, int> pii;
typedef pair<string, int> psi;
typedef pair<LL, LL> pll;
typedef map<string, int> msi;
typedef vector<int> vi;
typedef vector<LL> vl;
typedef vector<vl> vvl;
typedef vector<bool> vb;
inline bool scan_d(LL &num) {
    char in;bool IsN=false;
    in=getchar();
    if(in==EOF) return false;
    while(in!='-'&&(in<'0'||in>'9')) in=getchar();
    if(in=='-'){ IsN=true;num=0;}
    else num=in-'0';
    while(in=getchar(),in>='0'&&in<='9'){
            num*=10,num+=in-'0';
    }
    if(IsN) num=-num;
    return true;
}

const double eps = 1e-9;
const int maxn = 1010;
int n, k, m;
LL a[maxn], b[maxn];
pair<double, int> d[maxn];

bool ok(double x) {
    For(i, 1, n+1) d[i].first = a[i] - x * b[i], d[i].second = i;
    sort(d+1, d+n+1, greater<pair<double, int> >());
    double ta = .0, tb = .0;
    For(i, 1, n-k+1) {
        ta += a[d[i].second];
        tb += b[d[i].second];
    }
    return ta / tb >= x;
}

int main() {
    // FRead();
    while(~Rint(n) && ~Rint(k) && n + k) {
        For(i, 1, n+1) scan_d(a[i]);
        For(i, 1, n+1) scan_d(b[i]);
        double lo = 0, hi = 1;
        while(hi - lo > eps) {
            double mid = (lo + hi) / 2;
            if(ok(mid)) lo = mid;
            else hi = mid;
        }
        cout << fixed << setprecision(0) << lo * 100 << endl;
    }
    RT 0;
}