[poj 2976]Dropping tests 01分数规划

Dropping tests

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 5992 Accepted: 2076

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 15 0 25 1 64 21 2 7 95 6 7 90 0

Sample Output

83100

题目大意

手中有A场考试，每场考试答对的题目数为a[i],题目数为b[i],可以不计其中m场的结果，求答对题数/总题数的最大值

解题思路

01分数规划入门，算法来自http://www.cnblogs.com/perseawe/archive/2012/05/03/01fsgh.html

这个时候我们只需设d[i]=a[i]-b[i]*x，对于每一个x求sum{d[i]}是否大于0。

因为最优（最大）可行解只有一个，二分求解。

每次判断时只需要拿出d[i]最大的n-m个值即可。

code:

#include <cstdio>#include <cstring>#include <algorithm>#include <cmath> #include <string>#define eps 1e-9using namespace std;int cmp(double x,double y){return x>y;}double a[1005];double b[1005];double d[1005];int n,m,k;int main(){while(~scanf("%d%d",&n,&m),n+m){for (int i=1;i<=n;i++)scanf("%lf",&a[i]);for (int i=1;i<=n;i++)scanf("%lf",&b[i]);double ma=1,mi=0;while (ma-mi>eps){double mid=(ma+mi)/2;for (int i=1;i<=n;i++)d[i]=a[i]-mid*b[i];sort(d+1,d+1+n,cmp);double check=0;for (int i=1;i<=n-m;i++)check+=d[i];if (check>=0) mi=mid;else ma=mid;}printf("%.0f\n",ma*100);}return 0;}