Poj 2976 Dropping tests（01分数规划 牛顿迭代）

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
",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
",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;
}