poj 2976 Dropping tests

Dropping tests

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

Time Limit: 1000MS		Memory Limit: 65536K

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：n种数，每种数有2个属性，a，b ，选出n-k种数，最大化∑A（n-k）/∑B（n-k）

01分数规划模板题

转换：令∑Ak/∑Bk=ans ，最大化ans

假设只有2个数

（A1+A2）/(B1+B2)=ans

转化：A1+A2=(B1+B2)*ans

去括号、移项：A1-B1*ans+A2-B2*ans=0

假设指定一个f，设ans为最终答案

若f<ans,那么式子>0

若f>ans，那么式子<0

所以二分ans，每次取出前k个最大（Ai-Bi*ans）判断是否>0

若>0,移动下界，否则，移动上界

#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
using namespace std;
int n,k;
double a[1001],b[1001],tmp[1001];
double l,r,mid,p,ans;
int main()
{
    while(scanf("%d%d",&n,&k)!=EOF)
    {
        if(!n) break;
        for(int i=1;i<=n;i++) scanf("%lf",&a[i]);
        for(int i=1;i<=n;i++) scanf("%lf",&b[i]);
        l=0;r=1;
        while(fabs(l-r)>0.0001)
        {
            mid=(l+r)/2;p=0;
            for(int i=1;i<=n;i++) tmp[i]=a[i]-mid*b[i];
            sort(tmp+1,tmp+n+1);
            for(int i=n;i>k;i--) p+=tmp[i];
            if(p>0) l=mid;
            else r=mid;
        }
        printf("%.0lf\n",l*100);
    }
}

然而换了换二分姿势就错了，错误1：ans二分前ans要更新为0，防止不能二分

                                    错误2:0.0001精度太小，卡到0.0000001就过了

至于为啥上面0.0001就能A，玄学

以后想着精度卡6、7位就好

错误代码：（就是用新的变量ans，当式子>0时，更新ans，最后输出ans）

#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
using namespace std;
int n,k;
double a[1001],b[1001],tmp[1001];
double l,r,mid,p,ans;
int main()
{
    while(scanf("%d%d",&n,&k)!=EOF)
    {
        if(!n) break;
        for(int i=1;i<=n;i++) scanf("%lf",&a[i]);
        for(int i=1;i<=n;i++) scanf("%lf",&b[i]);
        l=0;r=1;
        while(fabs(l-r)>0.0001)
        {
            mid=(l+r)/2;p=0;
            for(int i=1;i<=n;i++) tmp[i]=a[i]-mid*b[i];
            sort(tmp+1,tmp+n+1);
            for(int i=n;i>k;i--) p+=tmp[i];
            if(p>=0) {ans=mid;l=mid+0.0001;}
            else r=mid-0.0001;
        }
        printf("%.0lf\n",ans*100);
    }
}