POJ 2976 Dropping tests(最大化平均值)

最新推荐文章于 2019-03-15 10:31:04 发布

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最新推荐文章于 2019-03-15 10:31:04 发布

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通过删除特定数量的考试成绩，计算并展示如何使考试平均分达到最大化的策略。

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题意：给出n门课程，并给出每门课的所考的成绩和这门课的总成绩，可以去掉k门课，求最大的平均分。

解题思路：平均分是sum(a[i]) / sum(b[i])，可行性函数判断的是， sum(a[i]) / sum(b[i]) >= x, 数学变换得sum(a[i] - x * b[i]) >= 0，所以每次进行可行性判断时先对a[i] - x * b[i]进行排序。

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).

#include<algorithm>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<cctype>
#include<list>
#include<iostream>
#include<map>
#include<queue>
#include<set>
#include<stack>
#include<vector>

using namespace std;

#define FOR(i, s, t) for(int i = (s) ; i <= (t) ; ++i)
#define REP(i, n) for(int i = 0 ; i < (n) ; ++i)

int buf[10];
inline long long read()
{
    long long x=0,f=1;
    char ch=getchar();
    while(ch<'0'||ch>'9')
    {
        if(ch=='-')f=-1;
        ch=getchar();
    }
    while(ch>='0'&&ch<='9')
    {
        x=x*10+ch-'0';
        ch=getchar();
    }
    return x*f;
}

inline void writenum(int i)
{
    int p = 0;
    if(i == 0) p++;
    else while(i)
        {
            buf[p++] = i % 10;
            i /= 10;
        }
    for(int j = p - 1 ; j >= 0 ; --j) putchar('0' + buf[j]);
}
/**************************************************************/
#define MAX_N 1005
const int INF = 0x3f3f3f3f;
int a[MAX_N];
int b[MAX_N];
double c[MAX_N];
int n, k;

bool f(double x)
{
//    cout<<1<<endl;
    for(int i = 0 ; i < n ; i++)
    {
        c[i] = a[i] - x * b[i];
    }
    sort(c, c + n);
//    for(int i = 0 ; i < n ; i++)
//    {
//        cout<<c[i]<<" ";
//    }
//    cout<<endl;
    double sum = 0;
    for(int i = 0 ; i < n - k ; i++)
    {
        sum += c[n - i - 1];
//        cout<<sum<<endl;
    }
//    cout<<sum<<endl;
    return sum >= 0;
}

int main()
{
//    freopen("in.txt", "r", stdin);
//	freopen("out.txt", "w", stdout);
    while(scanf("%d%d", &n, &k) && (n || k))
    {
        for(int i = 0 ; i < n ; i++)
        {
            a[i] = read();
        }
        for(int i = 0 ; i < n ; i++)
        {
            b[i] = read();
        }
        double lb = 0, ub = 1;
        while(ub - lb > 1e-8)
        {
            double mid = (lb + ub) / 2;
            if(f(mid)) lb = mid;
            else ub = mid;
        }
        int ans = floor(ub * 100 + 0.5);
        printf("%d\n", ans);
    }

    return 0;
}