POJ2976 Dropping tests (01分数规划)

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 npositive 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

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：问去掉k个数后∑a[i]/∑b[i]最大。

题解：01分数规划，二分答案就ok了。

 1 #include<cstdio>
 2 #include<algorithm>
 3 #include<iostream>
 4 #include<cstring>
 5 #include<cmath>
 6 #define N 1007
 7 #define eps 0.000001
 8 using namespace std;
 9 
10 int n,k;
11 double a[N],b[N],t[N];
12 
13 int main()
14 {
15     while(~scanf("%d%d",&n,&k)&&(n+k))
16     {
17         for (int i=1;i<=n;i++) scanf("%lf",&a[i]);
18         for (int i=1;i<=n;i++) scanf("%lf",&b[i]);
19         double l=0.0,r=10.0;
20         while(r-l>=eps)
21         {
22             double mid=(l+r)/2;
23             for (int i=1;i<=n;i++)
24                 t[i]=a[i]-mid*b[i];
25             sort(t+1,t+n+1);
26             double sum=0;
27             for (int i=k+1;i<=n;i++)
28                 sum+=t[i];
29             if (sum>=0) l=mid;
30             else r=mid;    
31         }
32         printf("%.0f
",l*100);
33     }
34 }