二分（01分数规划）

http://poj.org/problem?id=2976

题意：给出有n场考试，给出每场答对的题数a和这场一共有几道题b，求去掉k场考试后，公式.的最大值

解法：假设 r = sigma(ai*xi)/sigma(bi*xi) , 设R为最优解。变形后即有  sigma(ai*xi) - sigma(r*bi*xi) >= 0 . => sigma((ai-r*bi)*xi) >= 0 .

当 sigma((ai-r*bi)*xi) =0 时， 有R = r。 由因为 D[i] = ai-r*bi 为单调递减函数，二分 r 使 sigma((ai-r*bi)*xi) 逼近0 时 ， 由最大值R。

求出D数组后从大到小排列，从先前向后取N-K个即可，这时的D一定是最大的.

注意：这题要四舍五入，不能转int向下取整。

//#include<bits/stdc++.h>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <iostream>
#include <string>
#include <stdio.h>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <string.h>
#include <vector>
typedef long long ll ;
#define int ll
#define mod 1000000007
#define gcd __gcd
#define rep(i , j , n) for(int i = j ; i <= n ; i++)
#define red(i , n , j)  for(int i = n ; i >= j ; i--)
#define ME(x , y) memset(x , y , sizeof(x))
//ll lcm(ll a , ll b){return a*b/gcd(a,b);}
//ll quickpow(ll a , ll b){ll ans=1;while(b){if(b&1)ans=ans*a%mod;b>>=1,a=a*a%mod;}return ans;}
//int euler1(int x){int ans=x;for(int i=2;i*i<=x;i++)if(x%i==0){ans-=ans/i;while(x%i==0)x/=i;}if(x>1)ans-=ans/x;return ans;}
//const int N = 1e7+9; int vis[n],prime[n],phi[N];int euler2(int n){ME(vis,true);int len=1;rep(i,2,n){if(vis[i]){prime[len++]=i,phi[i]=i-1;}for(int j=1;j<len&&prime[j]*i<=n;j++){vis[i*prime[j]]=0;if(i%prime[j]==0){phi[i*prime[j]]=phi[i]*prime[j];break;}else{phi[i*prime[j]]=phi[i]*phi[prime[j]];}}}return len}
#define INF  0x3f3f3f3f
#define PI acos(-1)
#define pii pair<int,int>
#define fi first
#define se second
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
#define pb push_back
#define mp make_pair
#define cin(x) scanf("%lld" , &x);
using namespace std;
const int N = 1e7+9;
const int maxn = 1e5+9;
const double esp = 1e-2;
int n , k ;
double a[maxn] , b[maxn] , c[maxn];
bool check(double x){
    double ans = 0 ;
    rep(i , 1 , n){
        c[i] = a[i] - x * b[i];
    }
    sort(c + 1 , c + 1 + n);
    for(int i = n ; i > k ; i--){
        ans += c[i];
    }
    if(ans >= 0) return true;
    else return false;
}

void solve(){
    rep(i , 1 , n){
        scanf("%lf" , &a[i]);
    }
    rep(i , 1 , n){
        scanf("%lf" , &b[i]);
    }
    double l = 0 , r = INF;
    for(int i = 0 ; i < 100 ; i ++){
        double mid = (r + l) / 2 ;
        if(check(mid)){
            l = mid ;
        }else{
            r = mid ;
        }
    }
    printf("%.0f
" , l*100);
}

signed main()
{
    //ios::sync_with_stdio(false);
    //cin.tie(0); cout.tie(0);
    //int t ;
    //cin(t);
    //while(t--){
    while(~scanf("%lld%lld" , &n , &k) && (n || k))
        solve();
    //}
}