POJ 2976 Dropping tests(二分搜索，最大化平均值)


[align=center]Dropping tests[/align]

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 8233		Accepted: 2883

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).

题意：在一场测试中有N项，每一项都有bi个题目，答对ai个。总的答对率就是(a1+a2+....+an-1)/(b1+b2+....+bn-1)，现在可以让你从这N项测验中抽出K门不计入总的答题中，问最高答对率会是多少？

题解：二分搜索答对率，每得到一个值x，将其乘以bi，将yi=ai-bi*x，再将yi从大到小排序，看前n-k个yi的和是否大于等于0，是则表明当前x是最大答对率或小于最大答对率，否则表示当前x大于最大答对率。

代码如下：

#include<cstdio>
#include<cstring>
#include<algorithm>
#define INF 1000000010
using namespace std;
double a[1010],b[1010];
double y[1010];
int n,k;

int cmp(double a,double b)
{
return a>b;
}

bool dis(double x)
{
int i;
for(i=0;i<n;++i)
y[i]=a[i]-x*b[i];
sort(y,y+n,cmp);
double sum=0;
for(i=0;i<(n-k);++i)//注意是抽出k门课成绩不统计
sum+=y[i];
return sum>=0;
}

int main()
{
int i;
while(scanf("%d%d",&n,&k)&&n!=0||k!=0)
{
for(i=0;i<n;++i)
scanf("%lf",&a[i]);
for(i=0;i<n;++i)
scanf("%lf",&b[i]);
i=100;
double left=0,right=INF,mid;
while(i--)
{
mid=(left+right)/2;
if(dis(mid))
left=mid;
else
right=mid;
}
mid*=100;
printf("%.0f\n",mid);
}
return 0;
}