POJ2976 Dropping tests —— 01分数规划 二分法

题目链接：http://poj.org/problem?id=2976

Dropping tests

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 13615		Accepted: 4780

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

 

 

 

题解：

 

 

 

 

代码如下：

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <cmath>
 #include <algorithm>
 #include <vector>
 #include <queue>
 #include <stack>
 #include <map>
 #include <string>
 #include <set>
 #define ms(a,b) memset((a),(b),sizeof((a)))
 using namespace std;
 typedef long long LL;
 const double EPS = 1e-;
 const int INF = 2e9;
 const LL LNF = 2e18;
 const int MAXN = 1e3+;

 int a[MAXN], b[MAXN];
 double d[MAXN];
 int n, k;

 bool test(double L)
 {
     for(int i = ; i<=n; i++)
         d[i] = 1.0*a[i] - L*b[i];
     sort(d+, d++n);
     double sum = ;
     for(int i = k+; i<=n; i++) //舍弃前k小的数
         sum += d[i];
     return sum>=;
 }

 int main()
 {
     while(scanf("%d%d", &n, &k) && (n||k))
     {
         for(int i = ; i<=n; i++)
             scanf("%d", &a[i]);
         for(int i = ; i<=n; i++)
             scanf("%d", &b[i]);

         double l = , r = 1.0;
         while(l+EPS<=r)
         {
             double mid = (l+r)/;
             if(test(mid))
                 l = mid + EPS;
             else
                 r = mid - EPS;
         }
         printf("%.0f\n", r*);
     }
 }