Dropping tests - POJ 2976 二分-优快云博客

本文链接：https://blog.youkuaiyun.com/u014733623/article/details/27531479

本文探讨了一种方法，在特定课程中通过合理选择去除若干次测试成绩，来最大化累计平均分。详细解释了算法逻辑和实现步骤，并通过实例展示了如何计算最优平均分。

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Dropping tests

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 5774		Accepted: 2001

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).

题意：问n对数中去掉k对后的比最大平均数是多少。

思路：这道题用贪心做找最大价值的做法是不对的。二分枚举这个平均数，然后看如果最大平均数是这个，那么sum的和是否大于0，（sum的意思具体见代码吧）。此外这道题在挑战程序设计竞赛中的145页也有代码。

AC代码如下：

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
int n,k;
int w[1010];
int v[1010];
double y[1010];
bool solve(double x)
{ int i;
  double sum=0;
  for(i=1;i<=n;i++)
   y[i]=v[i]-w[i]*x;
  sort(y+1,y+1+n);
  for(i=1;i<=n-k;i++)
   sum+=y[n-i+1];
  if(sum>=0)
   return true;
  else
   return false;
}
int main()
{ int i,j;
  while(~scanf("%d%d",&n,&k) && n>0)
  { double l=0,r=1000000010,mi;
    for(i=1;i<=n;i++)
     scanf("%d",&v[i]);
    for(i=1;i<=n;i++)
     scanf("%d",&w[i]);
    for(i=1;i<=300;i++)
    { mi=(l+r)/2;
      if(solve(mi))
       l=mi;
      else
       r=mi;
    }
    printf("%.f\n",l*100);
  }
}