题目链接：http://poj.org/problem?id=2976
Dropping tests
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 17838 | Accepted: 6186 |
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
题目概述：
N个物品，每个物品 i 有 a[ i ] 和 b[ i ] 两种属性，去掉K个物品，是的 Σai / Σbi 最大。
解题思路：
我们令 Σai / Σbi >= x, 然后二分x，而判断条件则移一下项得 Σai >= Σbi*x → Σai - Σbi*x >= 0，因为我们要去掉 K 个小的， 取N-K个使结果最大化的，所以用 令 p[ i ] = Σai - Σbi*x，对 p[ i ] 默认升序排序之后，第 K 到 N 个。判断这些数和是否满足不等式条件（大于等于0）
注意事项：
因为计算机的精度问题，要注意细节的处理，对浮点数的计算。（in代码）
AC code：
#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cmath>
#define INF 0x3f3f3f3f
#define ll long long int
using namespace std; const int MAXN = ;
int N, K;
int a[MAXN], b[MAXN];
double p[MAXN]; ///这里的p一定要是浮点型，因为本来就是计算分式 bool ok(double x)
{
for(int i = ; i <= N; i++)
{
p[i] = a[i]-b[i]*x;
}
sort(p+, p+N+);
double ans = ;
for(int j = N; j > K; j--)
{
ans+=p[j];
}
return ans>=;
} int main()
{
while(~scanf("%d%d", &N, &K) && N|K)
{
for(int i = ; i <= N; i++)
{
scanf("%d", &a[i]);
}
for(int i = ; i <= N; i++)
{
scanf("%d", &b[i]);
}
double l = , r = ;
while(r-l>1e-) //!!!这里不用0，是因为计算机精度问题，计算浮点数要保留一定误差，改成0会超时
{
double mid = (l+r)/2.0;
if(ok(mid)) l = mid;
else r = mid;
}
printf("%.0lf\n", l*);
}
return ;
}