【poj 2976】 Dropping tests 二分（分数优化）

题目：http://poj.org/problem?id=2976

Dropping tests

Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 9121 Accepted: 3194

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

题意：最大化平均值

思路：贪心是错的，分数优化即可；
判断时a_i – x * b_i排序前k》=0————》true

代码：

#include
#include
#include
#include
#include
using namespace std;

int n,k;
double a[1005],b[1005];
double y[1005];
double cmp(double a,double b)
{
    return a>b;
}
bool C(double x)
{

    double tmp=0;
    for(int i=1;i<=n;i++)
    {
        y[i]=a[i]-x*b[i];
    }
    sort(y+1,y+1+n,cmp);
    for(int i=1;i<=k;i++)
    {
        tmp+=y[i];
    } 

    if(tmp>=0) return 1;
    return 0;

} 
void solve()
{
    double l=0,r=1.0,mid;
    for(int i=0;i<100;i++)
    {
        mid=(l+r)/2;
        if(C(mid))
        {
            l=mid;
        }else  r=mid;
    }

    printf("%.0lf\n",(double)mid*100.0);
}

int main()
{
    while(scanf("%d%d",&n,&k)!=EOF)
    {
        if(!n&&!k) return 0;
        for(int i=1;i<=n;i++)
        {
            scanf("%lf",&a[i]);
        }

        for(int j=1;j<=n;j++)
        scanf("%lf",&b[j]);
        k=n-k;
        solve();
    }
    return 0;
}