Dropping tests - POJ 2976 二分

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 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

.

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 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.

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).

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);
  }
}