[poj P2976] Dropping tests

最新推荐文章于 2023-09-03 14:09:56 发布

转载最新推荐文章于 2023-09-03 14:09:56 发布 · 153 阅读

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原文链接：http://www.cnblogs.com/whc200305/p/7750209.html

本文介绍了一种使用分数规划解决POJ P2976 Dropping Tests问题的方法，通过枚举和二分查找确定在丢弃k次测试成绩后能达到的最高累积平均分。

[poj P2976] Dropping tests

Time Limit: 1000MS Memory Limit: 65536K

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
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

Stanford Local 2005

01分数规划入门题。

这个不等式会经常看到：sigma(ai)/sigma(bi)>=(或<=)k

对于这题来说就是，找出最大的k，使得100*sigma(ai)/sigma(bi)>=k。

我们尝试化简上式。

因为bi>=0，所以sigma(bi)>=0，100*sigma(ai)>=k*sigma(bi)

100*sigma(ai)-k*sigma(bi)>=0

sigma(100*ai-k*bi)>=0

那么，我们只要先枚举一个k，将100*ai-k*bi作为关键字排序，再选出前n-k大的，判断一下sum是否非负就行了。

然后我们发现，上式满足单调性，枚举可以改为用二分。所以总复杂度是O(nlog分数)。

然后这一题我先开大100倍，然后最后再缩小100倍，但是这也会有精度误差。

总之，开大的倍数越大，误差越小，但效率越低，但也低不到哪里去qwq

code：

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<algorithm>
 4 #define LL long long
 5 #define M ((L)+(R)>>1)
 6 using namespace std;
 7 const int N=1005,lim=10000;
 8 int n,k,L,R,ans,final; LL sum;
 9 struct ob {LL a,b,c;}o[N];
10 bool cmp(ob x,ob y) {return x.c>y.c;}
11 bool jug(LL lv) {
12     for (int i=1; i<=n; i++) o[i].c=o[i].a*lim-lv*o[i].b;
13     sort(o+1,o+1+n,cmp);
14     sum=0;
15     for (int i=1; i<=n-k; i++) sum+=o[i].c;
16     return sum>=0;
17 }
18 int main() {
19     while (scanf("%d%d",&n,&k)!=EOF,n|k) {
20         for (int i=1; i<=n; i++) scanf("%lld",&o[i].a);
21         for (int i=1; i<=n; i++) scanf("%lld",&o[i].b);
22         for (L=0,R=lim; L<=R; )
23             if (jug(M)) ans=M,L=M+1;
24             else R=M-1;
25         final=ans/100; ans%=100;
26         if (ans>=50) final+=1;
27         printf("%d\n",final);
28     }
29     return 0;
30 }

View Code

转载于:https://www.cnblogs.com/whc200305/p/7750209.html