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).
##题意：
n场考试中分别答对a_i题，总题数分别为b_i，允许翘掉k场考试，求能达到的最高准确率。
##题解：
二分查找准确率即可。（这类题贪心肯定不行，通过样例也可以看出来）。
题目测试数据可以在这里找：http://ai.stanford.edu/~chuongdo/acm/2005/
#include<iostream>
#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;
const int maxn=1e5+5;
const double INF=1e6+5,esp=1e-6;
int v[maxn],w[maxn];
double y[maxn];
int n,k;
bool check(double x)//可以选择使得单位重量的价值不小于x
{
for(int i=0;i<n;i++)
{
y[i]=v[i]-x*w[i];
}
sort(y,y+n);
double sum=0;
for(int i=k;i<n;i++)
sum+=y[i];
return sum>=0;
}
void solve()
{
double lb=0,ub=INF;
while(lb+esp<ub)
{
double mid=(lb+ub)/2;
if(check(mid))
lb=mid;
else
ub=mid;
}
printf("%.f
",floor(100*lb+0.5));
}
int main()
{
while(cin>>n>>k,n||k)
{
for(int i=0;i<2*n;i++)
{
if(i<n)
scanf("%d",&v[i]);
else
scanf("%d",&w[i%n]);;
}
solve();
}
return 0;
}