HDU2976 Dropping tests 2017-05-11 18:10 39人阅读 评论(0) 收藏

Dropping tests

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 12187		Accepted: 4257

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

————————————————————————————————

给出n和分数，可以去其中m个，求剩下的分子之和除以分母之和最大

思路：二分答案，验证去了m个后能否达到

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <string>
#include <vector>
using namespace std;
#define inf 0x3f3f3f3f
#define LL long long

int n,m;
struct node{
LL a,b;
}p[100005];
double k;

bool cmp(node a,node b)
{
    double x=a.a-(a.b*k);
    double y=b.a-(b.b*k);
    return x>y;
}

bool ok(double mid)
{
    k=mid;
    sort(p,p+n,cmp);
    LL x=0,y=0;
    for(int i=0;i<n-m;i++)
    {
        x+=p[i].a;
        y+=p[i].b;
    }
    double ans=x*1.0/y;
    if(ans>=mid)
        return 1;
    return 0;

}

int main()
{
    while(~scanf("%d%d",&n,&m)&&(m||n))
    {
        for(int i=0;i<n;i++)
            scanf("%lld",&p[i].a);
            for(int i=0;i<n;i++)
            scanf("%lld",&p[i].b);

    double l=0,r=1;
    while(r-l>1e-5)
    {
    double mid=(l+r)/2;
        if(ok(mid))
        {
            l=mid;
        }
        else
            r=mid;
    }
    l*=100;
    printf("%.0f\n",l);
    }
    return 0;
}