POJ 2976/ZOJ 3068 Dropping tests 01分数规划

卡精度的01分数规划...

Dropping tests

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6464		Accepted: 2230

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

Stanford Local 2005

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>

using namespace std;

const int maxn=1111;
const double eps=1e-8;

int n,k;
double a[maxn],b[maxn];
double d[maxn];

bool check(double L)
{
for(int i=0;i<n;i++)
{
d[i]=a[i]-b[i]*L;
}
sort(d,d+n);
double sum=0.;
for(int i=k;i<n;i++)
{
sum+=d[i];
}
if(sum>eps)
return true;
return false;
}

int main()
{
while(scanf("%d%d",&n,&k)!=EOF)
{
if(n==0&&k==0) break;
for(int i=0;i<n;i++)
scanf("%lf",a+i);
for(int i=0;i<n;i++)
scanf("%lf",b+i);

double low=0,high=1,mid,ans;
while(low+eps<high)
{
mid=(low+high)/2.;
if(check(mid))
{
low=mid; //ans=mid;这样就WA了
}
else
high=mid;
}
//cout<<"ans: "<<ans<<endl;
printf("%.0f\n",100*low);
}
return 0;
}