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最近遇到了一类问题在翻阅大神博客后方知是01分数规划问题
感谢大神博客：http://blog.csdn.net/hhaile/article/details/8883652
解题思路：将公式化简F(L)=sigma(a[i]*x[i])-L*sigma(b[i]*x[i])，当确定一个L时，如果还有符合条件的L并且比原来的大，此时f》0，由于结果是1--100之内的数，所以最多需要找100次
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
#include <iostream>
#include <stdio.h>
#include <cstring>
#include <algorithm>
using namespace std;
int n,k;
int a[1002],b[1002];
double c[1002];
bool wo(double mid)
{
for(int i=0;i<n;i++)
c[i]=(double)a[i]-(double)b[i]*mid;
sort(c,c+n);
double sum=0;
for(int i=k;i<n;i++)
sum+=c[i];
return sum>0;
}
int main()
{
while(~scanf("%d%d",&n,&k))
{
if(n==0&&k==0)
break;
for(int i=0;i<n;i++)
scanf("%d",&a[i]);
for(int i=0;i<n;i++)
scanf("%d",&b[i]);
double l=0,r=1,mid;
for(int i=0;i<100;i++)
{
mid=(l+r)/2;
if(wo(mid))
l=mid;
else
r=mid;
}
printf("%d\n",(int)(100*mid+0.5));
}
return 0;
}
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