Dropping tests

题目链接：http://poj.org/problem?id=2976

Dropping tests

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 20830		Accepted: 7052

Description

In a certain course, you take n tests. If you get a_i out of b_i questions correct on test i, your cumulative average is defined to be

.

Given your test scores and a positive integer k, determine how high you can make your cumulative average if you are allowed to drop any k of your test scores.

Suppose you take 3 tests with scores of 5/5, 0/1, and 2/6. Without dropping any tests, your cumulative average is . However, if you drop the third test, your cumulative average becomes .

Input

The input test file will contain multiple test cases, each containing exactly three lines. The first line contains two integers, 1 ≤ n ≤ 1000 and 0 ≤ k < n. The second line contains n integers indicating a_i for all i. The third line contains npositive integers indicating b_i for all i. It is guaranteed that 0 ≤ a_i ≤ b_i ≤ 1, 000, 000, 000. The end-of-file is marked by a test case with n = k = 0 and should not be processed.

Output

For each test case, write a single line with the highest cumulative average possible after dropping k of the given test scores. The average should be rounded to the nearest integer.

Sample Input

3 1
5 0 2
5 1 6
4 2
1 2 7 9
5 6 7 9
0 0

Sample Output

83
100

Hint

To avoid ambiguities due to rounding errors, the judge tests have been constructed so that all answers are at least 0.001 away from a decision boundary (i.e., you can assume that the average is never 83.4997).

题目大意：输入N K  下面有两行 每行有N个数  第一行代表N个a[i]   第二行代表N个b[i]   问你按照上面的公式算  去掉K个  得到的最大结果是多少

思路：自己并没有想到怎么二分 ，自己想到的是怎么贪心，但是感觉不行，就没去尝试了，其实想的到二分答案，但是怎么判断答案是否成立就没有想到了，这里用到的一种思想，分析表达式

以前我都是看到表达式就看一下的，从来没有给它仔细化简分析啥的  这一题算是一个教训吧！  那么下面看一下这个表达式到底怎么回事

令r = ∑a[i] * x[i] / (b[i] * x[i])  则必然∑a[i] * x[i] - ∑b[i] * x[i] * r= 0；（条件1）并且任意的 ∑a[i] * x[i] - ∑b[i] * x[i] * max(r) <= 0  (条件2，只有当∑a[i] * x[i] / (b[i] * x[i]) = max(r) 条件2中等号才成立)然后就可以枚举r , 对枚举的r， 求Q(r) = ∑a[i] * x[i] - ∑b[i] * x[i] * r  的最大值，  为什么要求最大值呢？  因为我们之前知道了条件2，所以当我们枚举到r为max(r)的值时，显然对于所有的情况Q(r)都会小于等于0，并且Q(r)的最大值一定是0.而我们求最大值的目的就是寻找Q(r)=0的可能性，这样就满足了条件1，最后就是枚举使得Q(r)恰好等于0时就找到了max(r)。而如果能Q(r)>0 说明该r值是偏小的，并且可能存在Q(r)=0,而Q(r)<0的话，很明显是r值偏大的，因为max(r)都是使Q(r)最大值为0，说明不可能存在Q(r)=0了。
注意：二分真的很气人啊，这题用了以前经常用的二分板子，竟然错了。。。  真的不明白为啥，就是二分最后的结果是l  还是r   应该是这题有点问题，下面wa的代码也给出来吧

wa的：

#include<iostream>
#include<algorithm>
#include<stdio.h>
using namespace std;
typedef long long ll;
const int maxn=+;
const double inf=1e15;
const double eps=1e-;
int N,K;
ll a[maxn],b[maxn];
double c[maxn];
bool cmp(const double x,const double y)
{
    return x>y;
}
bool judge(double mid)
{
    for(int i=;i<N;i++)//找到最优的解
    {
        c[i]=a[i]-mid*b[i];
    }
    sort(c,c+N,cmp);
    double sum=;
    for(int i=;i<N-K;i++) sum+=c[i];
    return sum>=;
}
int main()
{
    while(cin>>N>>K)
    {
        if(N==&&K==) break;
        for(int i=;i<N;i++) cin>>a[i];
        for(int i=;i<N;i++) cin>>b[i];
        double l=,r=inf;
        double ans;

        while(r-l>=eps)
        {
            double mid=(r+l)/;
            //cout<<mid<<endl;
            if(judge(mid))
            {
                ans=mid;
                l=mid+eps;
            }
            else r=mid;
            //cout<<"ans: "<<ans<<endl;
            //cout<<"l: "<<l<<endl;
            //cout<<"r: "<<r<<endl;
       }
        printf("%.0lf\n", ans * );

/*
        double mid;
        while(r-l>=eps)
        {

            mid=(l+r)/2;
            if(judge(mid))
                l=mid;
            else r=mid;
            cout<<"l: "<<l<<endl;
            cout<<"r: "<<r<<endl;
        }
        printf("%.0lf\n",r*100);
        //printf("%.0lf\n", ans * 100);
        //cout<<(int)(ans*100+0.5)<<endl;
*/
    }
    return ;
}

/*
*/

ac的：

#include<iostream>
#include<algorithm>
#include<stdio.h>
using namespace std;
typedef long long ll;
const int maxn=+;
const double inf=1e15;
const double eps=1e-;
int N,K;
ll a[maxn],b[maxn];
double c[maxn];
bool cmp(const double x,const double y)
{
    return x>y;
}
bool judge(double mid)
{
    for(int i=;i<N;i++)//找到最优的解
    {
        c[i]=a[i]-mid*b[i];
    }
    sort(c,c+N,cmp);
    double sum=;
    for(int i=;i<N-K;i++) sum+=c[i];
    return sum>=;
}
int main()
{
    while(cin>>N>>K)
    {
        if(N==&&K==) break;
        for(int i=;i<N;i++) cin>>a[i];
        for(int i=;i<N;i++) cin>>b[i];
        double l=,r=inf;
        double ans;

        while(r-l>=eps)
        {
            double mid=(r+l)/;
            //cout<<mid<<endl;
            if(judge(mid))
            {
                ans=mid;
                l=mid+eps;
            }
            else r=mid;
            //cout<<"ans: "<<ans<<endl;
            //cout<<"l: "<<l<<endl;
            //cout<<"r: "<<r<<endl;
       }
        printf("%.0lf\n", r * );

/*
        double mid;
        while(r-l>=eps)
        {

            mid=(l+r)/2;
            if(judge(mid))
                l=mid;
            else r=mid;
            cout<<"l: "<<l<<endl;
            cout<<"r: "<<r<<endl;
        }
        printf("%.0lf\n",r*100);
        //printf("%.0lf\n", ans * 100);
        //cout<<(int)(ans*100+0.5)<<endl;
*/
    }
    return ;
}

/*
*/