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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 .
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.
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.
3 15 0 25 1 64 21 2 7 95 6 7 90 0
83100
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).
开始陷入了思维误区,想着贪心解题,按照分子与分母差值从小到大排,还以为很对。仔细检验一下发现差值小未必对整体贡献大,正解实际应当是01分数规划。我们需要求解Σa[i]/Σb[i]的最大值,不妨设此最大值为max,则可知Σa[i]==Σb[i]*max。利用二分,我们可以不断取得mid,并通过Σa[i]与Σb[i]*mid的实际情况判断其相对于标准答案的大小关系。
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