POJ 2976 -- Dropping tests
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
1 #include<cstdio>
2 #include<iostream>
3 #include<algorithm>
4 using namespace std;
5 #define Max 1008
6 #define eps 1e-4
7 bool cmp(double a,double b)
8 {
9 return a>b;
10 }
11 int main()
12 {
13 int n,k,i;
14 double left,mid,right,ans;
15 double score[Max],a[Max],b[Max];
16 while(scanf("%d %d",&n,&k),n+k)
17 {
18 for(i=0;i<n;i++) scanf("%lf",&a[i]);
19 for(i=0;i<n;i++) scanf("%lf",&b[i]);
20 for(left=0,right=100;right-left>eps;)
21 {
22 mid=(left+right)/2.0;
23 for(i=0;i<n;i++)
24 score[i]=a[i]*100.0-b[i]*mid;
25 sort(score,score+n,cmp);
26 ans=0;
27 for(i=0;i<n-k;i++)
28 ans+=score[i];
29 if(ans>0)
30 left=mid;
31 else
32 right=mid;
33 }
34 cout<<(int)(left+.5)<<endl;
35 }return 0;
36 }