POJ 2976 Dropping tests 【01分数规划+二分】-优快云博客

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 a_i out of b_i 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 a_i for all i. The third line contains n positive integers indicating b_i for all i. It is guaranteed that 0 ≤ a_i ≤ b_i ≤ 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

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

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：

 1 #include <cstdio>
 2 #include <iostream>
 3 #include <algorithm>
 4 #include <cstring>
 5 #include <cmath>
 6 #define INF 0x3f3f3f3f
 7 #define ll long long int
 8 using namespace std;
 9 
10 const int MAXN =  2048;
11 int N, K;
12 int a[MAXN], b[MAXN];
13 double p[MAXN];       ///这里的p一定要是浮点型，因为本来就是计算分式
14 
15 bool ok(double x)
16 {
17     for(int i = 1; i <= N; i++)
18     {
19         p[i] = a[i]-b[i]*x;
20     }
21     sort(p+1, p+N+1);
22     double ans = 0;
23     for(int j = N; j > K; j--)
24     {
25         ans+=p[j];
26     }
27     return ans>=0;
28 }
29 
30 int main()
31 {
32     while(~scanf("%d%d", &N, &K) && N|K)
33     {
34         for(int i = 1; i <= N; i++)
35         {
36             scanf("%d", &a[i]);
37         }
38         for(int i  = 1; i <= N; i++)
39         {
40             scanf("%d", &b[i]);
41         }
42         double l = 0, r = 1;
43         while(r-l>1e-6)        //!!!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
44         {
45             double mid = (l+r)/2.0;
46             if(ok(mid)) l = mid;
47             else r = mid;
48         }
49         printf("%.0lf\n", l*100);
50     }
51     return 0;
52 }