POJ2976————Dropping tests（二分法，最大化平均值）

Dropping tests

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 10267 Accepted: 3584

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 n positive 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 15 0 25 1 64 21 2 7 95 6 7 90 0

Sample Output

83100

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).

平均值为sigma（a）/sigma(b).

如果一个平均值可以大于一个数x

那么就等价于

sigma(a)/sigma(b)-x>=0;

转化一下子也就是sigma(a[i]-b[i]*x)>0

我们按照这个值排序，选择最大的那n-k个数判断一下即可

#include <cstdio>#include <algorithm>using namespace std;const int MAXN =1010;const int INF=1000000007;int a[MAXN],b[MAXN];double s[MAXN];int cmp(double x,double y){return x>y;}int n,k;bool C(double x){for(int i=0;i<n;i++){s[i]=a[i]-x*b[i];}sort(s,s+n,cmp);double sum=0;for(int i=0;i<n-k;i++){sum+=s[i];}return sum>=0;}int main(){while(scanf("%d%d",&n,&k)){    if(n==0) break;for(int i=0;i<n;i++){scanf("%d",a+i);}for(int i=0;i<n;i++)scanf("%d",b+i);double lb=0,ub=INF;for(int i=0;i<100;i++){double m=(lb+ub)/2;if(C(m))lb=m;elseub=m;}lb*=100;int res;if(lb-int(lb)<(int)lb+1-lb){res=(int)lb;}elseres=(int)lb+1;printf("%d\n",res);}}