POJ——T 2976 Dropping tests

http://poj.org/problem?id=2976

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 13861		Accepted: 4855

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 npositive 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

 

 

题意：给出n个物品，每个物品有两个属性a和b，选择n-k个元素，询问sum{ai}/sum{bi}的最大值。

分数规划 二分答案ans ， 判断 sum[a[i]]/sum[b[i]]与ans的关系

　　即 判断 b[i]*ans-a[i]*100+b[i+1]*ans-a[i+1]*100+...+b[i+k]*ans-a[i+k]*100<0

#include <algorithm>
#include <cstdio>

inline void read(int &x)
{
    x=0; register char ch=getchar();
    for(; ch>'9'||ch<'0'; ) ch=getchar();
    for(; ch>='0'&&ch<='9'; ch=getchar()) x=x*10+ch-'0';
}
const double eps(1e-9);
const int N(1005);
double a[N],b[N];
int n,k;

double l,r,mid,ans,tmp[N];
inline bool check(double x)
{
    double sum=0.0;
    for(int i=1; i<=n; ++i)
        tmp[i]=1.0*b[i]*x-100.0*a[i];
    std::sort(tmp+1,tmp+n+1);
    for(int i=1; i<=n-k; ++i) sum+=tmp[i];
    return sum<0;
}

int Presist()
{
    for(; 1; )
    {
        read(n),read(k);        if(!n&&!k) break;
        for(int i=1; i<=n; ++i) scanf("%lf",&a[i]);
        for(int i=1; i<=n; ++i) scanf("%lf",&b[i]);
        for(l=0,r=100.0; r-l>eps; )
        {
            mid=(l+r)/2.0;
            if(check(mid))
                 l=mid;
            else r=mid;
        }
        printf("%.0lf
",l);
    }
    return 0;
}

int Aptal=Presist();
int main(int argc,char**argv){;}