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 npositive integers indicating bi for all i. It is guaranteed that 0 ≤ ai ≤ bi ≤ 1,000,000. 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).
#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <queue> #include <cmath> using namespace std; #define eps 1e-7 int n,k; double a[2005],b[2005],t[2005]; double gh(double x) { for(int i=0;i<n;i++) t[i]=a[i]-x*b[i]; sort(t,t+n); double ans=0; for(int i=k;i<n;i++) ans+=t[i]; return ans; } int main() { while(scanf("%d %d",&n,&k)) { if(n==0&&k==0) break; for(int i=0;i<n;i++) scanf("%lf",&a[i]); for(int i=0;i<n;i++) scanf("%lf",&b[i]); double l=0.0,r=1.0,mid; while(r-l>eps) { mid=(l+r)/2; if(gh(mid)>0) l=mid; else r=mid; } printf("%1.f\n",l*100); } return 0; }
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