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

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=1000+5;
const double inf=1e15;
const double eps=1e-7;
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=0;i<N;i++)//找到最优的解
    {
        c[i]=a[i]-mid*b[i];
    }
    sort(c,c+N,cmp);
    double sum=0;
    for(int i=0;i<N-K;i++) sum+=c[i];
    return sum>=0;
}
int main()
{
    while(cin>>N>>K)
    {
        if(N==0&&K==0) break;
        for(int i=0;i<N;i++) cin>>a[i];
        for(int i=0;i<N;i++) cin>>b[i];
        double l=0,r=inf;
        double ans;

        while(r-l>=eps)
        {
            double mid=(r+l)/2;
            //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 * 100);

/*
        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 0;
}

/*
*/

ac的：

#include<iostream>
#include<algorithm>
#include<stdio.h>
using namespace std;
typedef long long ll;
const int maxn=1000+5;
const double inf=1e15;
const double eps=1e-7;
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=0;i<N;i++)//找到最优的解
    {
        c[i]=a[i]-mid*b[i];
    }
    sort(c,c+N,cmp);
    double sum=0;
    for(int i=0;i<N-K;i++) sum+=c[i];
    return sum>=0;
}
int main()
{
    while(cin>>N>>K)
    {
        if(N==0&&K==0) break;
        for(int i=0;i<N;i++) cin>>a[i];
        for(int i=0;i<N;i++) cin>>b[i];
        double l=0,r=inf;
        double ans;

        while(r-l>=eps)
        {
            double mid=(r+l)/2;
            //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 * 100);

/*
        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 0;
}

/*
*/

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