hdu Dropping tests 0/1分数规划（二分求值）

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 .

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.

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.

y=100*sigma(ai)/sigma(bi)

t(y)=100*sigma(ai)-y*sigma(bi)

问最后去掉那几个k能得到正好的值，

***************************************************************************************************************************

 1 #include<iostream>
 2 #include<string>
 3 #include<cstring>
 4 #include<cstdio>
 5 #include<cmath>
 6 #include<queue>
 7 #include<algorithm>
 8 using namespace std;
 9 double a[1001],b[1001],score[1001];
10 int n,k,i,j;
11 const double eps=1e-8;
12 int main()
13 {
14     while(scanf("%d %d",&n,&k)&&(n+k))
15     {
16         for(i=0;i<n;i++)
17             scanf("%lf",&a[i]);
18         for(i=0;i<n;i++)
19             scanf("%lf",&b[i]);
20         double le=0,ri=100,mid,sum;
21         while(ri-le>eps)
22         {
23             mid=(ri+le)/2;
24             for(i=0;i<n;i++)
25                 score[i]=100*a[i]-b[i]*mid;
26             sort(score,score+n);
27             sum=0;
28             for(i=k;i<n;i++)
29                 sum+=score[i];
30             if(sum>0)
31                 le=mid;
32             else
33                 ri=mid;
34         }
35         printf("%.0f\n",le);
36     }
37     return 0;
38 }

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