Dropping tests（二分查找平均值）

原题链接
Dropping tests
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 9915 Accepted: 3459
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

这个问题在《挑战程序设计竞赛》上有完全类似的问题就是求最大化平均值，这类问题可以归结为一类问题那就是有n个物品的重量和价值分别是wi和vi，从中选出k个物品使得单位重量的价值最大。
那么首先我们可以规定C(x)的意思为可以选择使得单位重量的价值不小于x。假设选了k个物品。那么必有这k个物品有如下图的关系

要得到这个k个值所以这个时候只需要把进行了v-x*w运算后的值进行从大到小的排序后选择前k个即可

//http://poj.org/problem?id=2976#include <algorithm>#include <iostream>#include <utility>#include <sstream>#include <cstring>#include <cstdio>#include <vector>#include <queue>#include <stack>#include <cmath>#include <map>#include <set>using namespace std;typedef long long ll;const int MOD = int(1e9) + 7;//int MOD = 99990001;const int INF = 0x3f3f3f3f;const ll INFF = (~(0ULL)>>1);const double EPS = 1e-9;const double OO = 1e20;const double PI = acos(-1.0); //M_PI;const int fx[] = {-1, 1, 0, 0};const int fy[] = {0, 0, -1, 1};const int maxn=1000 + 5;struct node{int u,d;}a[maxn];int n,k,MAX;double shengyu[maxn];//判断下当前的这个平均数是否可以bool cmp(double x,double y){        return x>y;}bool can(double mid){        for(int i=0;i<n;i++)                shengyu[i]=a[i].u - mid*a[i].d;        sort(shengyu,shengyu+n,cmp);        //计算从大到小前k项的和        double sum=0;        for(int i=0;i<(n-k);i++)                sum+=shengyu[i];        return sum>=0;}void solve(){        double l=0,r=1;        for(int i=0;i<100;i++){                double mid=(l+r)/2;                //cout << "mid:" << mid << endl;                if(can(mid)) l=mid;                else r=mid;        }        printf("%d\n",(int)(r*100 + 0.5));}int main(){        while(scanf("%d%d",&n,&k)==2 && (n+k)){                //k=n-k;//把题目修改成选择k个科目使得平均分最大，并且这个时候最少还会选一门                for(int i=0;i<n;i++)                        scanf("%d",&a[i].u);                for(int i=0;i<n;i++)                        scanf("%d",&a[i].d);                solve();        }        return 0;}