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-8;
const int INF = 2e9;
const LL LNF = 2e18;
const int MAXN = 1e3+10;

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

bool test(double L)
{
    for(int i = 1; i<=n; i++)
        d[i] = 1.0*a[i] - L*b[i];
    sort(d+1, d+1+n);
    double sum = 0;
    for(int i = k+1; i<=n; i++) //舍弃前k小的数
        sum += d[i];
    return sum>=0;
}

int main()
{
    while(scanf("%d%d", &n, &k) && (n||k))
    {
        for(int i = 1; i<=n; i++)
            scanf("%d", &a[i]);
        for(int i = 1; i<=n; i++)
            scanf("%d", &b[i]);

        double l = 0, r = 1.0;
        while(l+EPS<=r)
        {
            double mid = (l+r)/2;
            if(test(mid))
                l = mid + EPS;
            else
                r = mid - EPS;
        }
        printf("%.0f\n", r*100);
    }
}