Dropping tests

Time Limit: 1000MS Memory Limit: 65536K

Description

In a certain course, you take n tests. If you get ai out of bi 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 ai for all i. The third line contains n positive integers indicating bi for all i. It is guaranteed that 0 ≤ ai ≤ bi ≤ 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).

Source

Stanford Local 2005

/*

裸的01分数规划问题.

令∑a[i]/∑b[i]=ans.

则∑a[i]-∑b[i]*ans=0.

二分一个ans.

然后用a[i]-b[i]*ans取前k大检验.

只能去感性的认识orz...

并不会证明..

*/

#include<iostream>

#include<cstdio>

#include<algorithm>

#define eps 1e-7

#define MAXN 1001

using namespace std;

double ans,a[MAXN],b[MAXN],sum,tmp[MAXN];

int n,m,k;

bool check(double x)

{

    double tot=0;

    for(int i=1;i<=n;i++) tmp[i]=a[i]-x*b[i];

    sort(tmp+1,tmp+n+1,greater<double>());

    for(int i=1;i<=n-k;i++) tot+=tmp[i];

    if(tot>=0) return true;

    else return false;

}

void slove()

{

    double l=0,r=1e4,mid;

    while(l<=r)

    {

        mid=(l+r)/2.0;

        if(check(mid)) l=mid+eps,ans=mid;

        else r=mid-eps;

    }

    printf("%.0f\n",ans*100);

    return ;

}

int main()

{

    while(scanf("%d%d",&n,&k))

    {

        if(!n&&!k) break;

        sum=ans=0;

        for(int i=1;i<=n;i++) scanf("%lf",&a[i]);

        for(int i=1;i<=n;i++) scanf("%lf",&b[i]);

        slove();

    }

    return 0;

}

/*

发现这题牛顿迭代可做吖.

网上的题解都是二分01规划的.

我就写个牛顿迭代的吧orz(虽然二分的写过).

先选一个估计值s0.

我们能保证这个答案是单调的.

假设上次迭代的ans为s1,

则存在n-k个元素使s1=∑(ai/bi),

变形可得到∑ai-s2*∑bi=0,

令ans[i]=a[i]-b[i]*s0.

取前n-k大统计一个答案.

可知必存在n-k个元素使∑ansi=∑ai-s1*∑bi=0,

所以当我们按ans排序并取前n-k个元素作为求其∑ans时,

∑ansi显然是>=0的,

然后s1=(∑ai-∑ansi)/∑bi)<=(∑ai/∑bi)=s2(i<=n-k).

即此迭代过程是收敛的,当等号成立时,s即为答案.

有些地方还是有点想不通毕竟弱吖orz.

*/

#include<cstdio>

#include<algorithm>

#include<cmath>

#define MAXN 1001

#define eps 1e-7

using namespace std;

double ans,sum,tmp[MAXN];

int n,m,k;

struct data{double a,b,ans;}s[MAXN];

bool cmp(const data &x,const data &y)

{

    return x.ans>y.ans;

}

void slove()

{

    double suma=0,sumb=0,s0=0,s1=0;

    for(int i=1;i<=k;i++) suma+=s[i].a,sumb+=s[i].b;

    s0=suma/sumb;

    while(abs(s0-s1)>eps)

    {

        s1=s0;suma=sumb=0;

        for(int i=1;i<=n;i++) s[i].ans=s[i].a-s[i].b*s0;

        sort(s+1,s+n+1,cmp);

        for(int i=1;i<=k;i++) suma+=s[i].a,sumb+=s[i].b;

        s0=suma/sumb;

    }

    printf("%.0f\n",s0*100);

    return ;

}

int main()

{

    while(~scanf("%d%d",&n,&k))

    {

        if(!n&&!k) break;

        sum=ans=0;k=n-k;

        for(int i=1;i<=n;i++) scanf("%lf",&s[i].a);

        for(int i=1;i<=n;i++) scanf("%lf",&s[i].b);

        slove();

    }

    return 0;

}