d F - Dropping test.（ai+加和 bi+ n k 求最小分数）

加和二分

为啥这样算

n 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 1 5 0 2 5 1 6 4 2 1 2 7 9 5 6 7 9 0 0 Sample Output 83 100 Hint

#include <iostream> #include<algorithm> #include<functional> using namespace std; const int maxn=1000+5; const int INF=1e8+10; int n,k; double a[maxn],b[maxn],c[maxn]; bool judge(double mid) { for(int i=0;i<n;i++) c[i]=a[i]-mid*b[i];// sort(c,c+n,greater<double>()); double ans=0; for(int i=0;i<k;i++) ans+=c[i]; return ans>=0; } int main() { while(cin>>n>>k&&n) { k=n-k; for(int i=0;i<n;i++) { cin>>a[i]; a[i]=a[i]*100; } for(int i=0;i<n;i++) cin>>b[i]; double l=0,r=100.0; while(r-l>0.000001) { double mid=(l+r)/2.0; if(judge(mid)) l=mid; else r=mid; } int ans=(int )(l+0.5); cout << ans<< endl; } return 0; }