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2013
11-12

POJ 2976 Dropping tests [解题报告] Java

Dropping tests

问题描述 :

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 .

输入:

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 ≤ aibi ≤ 1, 000, 000, 000. The end-of-file is marked by a test case with n = k = 0 and should not be processed.

输出:

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.

样例输入:

3 1
5 0 2
5 1 6
4 2
1 2 7 9
5 6 7 9
0 0

样例输出:

83
100

温馨提示:

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).

解题代码:

//* @author: 
import java.util.Scanner;
import java.util.Arrays;
public class Main{
  public static void main(String args[]){
    Scanner sc=new Scanner(System.in);
    int i,j,n,k;
    while (sc.hasNext()) {
        n=sc.nextInt();
        k=sc.nextInt();
       if(n==0) break;
        int num[]=new int[n];
        int den[]=new int[n];
        double scores[]=new double[n];
        for (i=0;i< n;i++) 
            num[i]=sc.nextInt();
        for (i=0;i< n;i++)
            den[i]=sc.nextInt();
        double lb = 0, ub = 1;
        for (i = 0; i < 100; i++) {
            double x = (lb+ub)/2;
            for (j = 0; j < n; j++)
	        scores[j] = num[j] - x*den[j];
            Arrays.sort(scores);
            double total = 0;
            for (j = k; j < n; j++)
	        total += scores[j];
            if (total >= 0) lb = x;
            else ub = x;
         }
        System.out.printf("%d\n",(int)(100*lb + 0.5));
    }
  }
}