Dropping tests--计算方法，01分数规划
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 10257 | Accepted: 3578 |
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 npositive 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).
题目链接：http://poj.org/problem?id=2976
作为poj为数不多的好理解题意的题，题意很好理解，给了你一个式子，让你求他的最大，我就直接排了一下序，好吧，错了，我感觉不对，要考虑每一个数的贡献，但是一直不会考虑，以为排一下序就可以了，但是很明显，wa。网上查了一下，01分数规划经典题，好吧，继续扩充自己的知识面。
放上大牛的链接，本想自己写写题解，但是怎么写都觉的没有大牛好，那就直接把大牛的链接放这
http://www.cnblogs.com/perseawe/archive/2012/05/03/01fsgh.html
讲的很好
代码：
#include
#include
#include
#include
#define inf 0x3f3f3f3f
#define eps 1e-7
using namespace std;
double a[2000];
double b[2000];
double t[2000];
int n,m;
double fax(double L){
for(int i=0;ieps){
mid=(l+r)/2;
if(fax(mid)>0)
l=mid;
else
r=mid;
}
printf("%1.f\n",l*100);
}
return 0;
}