POJ2976基础01分数规划

In a certain course, you take n tests. If you get ai out of bi 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 ai for all i. The third line contains n positive integers indicating bi for all i. It is guaranteed that 0 ≤ ai ≤ bi ≤ 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).//只输出整数

01二分规划的意思是对于某个数，0，1表示选择与否

AC代码：

#include <iostream>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <cstdio>
#define eps 1e-8
#define maxn 10000
using namespace std;
int main()
{
    double a[maxn],b[maxn],c[maxn];//后面mid为double型的 ，最好把所有有可能的地方都换成double的
    int n,i,j,k;
    while(scanf("%d%d",&n,&k)&&(n!=0||k!=0))
    {
        for(i=0;i<n;i++)
            scanf("%lf",&a[i]);
        for(i=0;i<n;i++)
            scanf("%lf",&b[i]);
        double l=0,r=1000;//左右边界，r开的稍微大一点也没关系
        double mid;
        while(r-l>eps)//即对于浮点数的比较，r>l+eps,即r>l，所以break时r=l
        {
            mid=(r+l)/2;
            for(i=0;i<n;i++)
                c[i]=a[i]*100-mid*b[i];
            sort(c,c+n);
            double maxf=0;
            for(i=n-1;i>=k;i--)
                maxf+=c[I];//求最大，即比较删掉前面k个小的c[i]之后剩下数的和是否大于0，二分
            if(maxf>eps) l=mid;
            else r=mid;
        }
        printf("%.0lf\n",l);//此处可以输出l，r，mid
    }
    return 0;
}