【POJ 2976 Dropping tests】

Time Limit: 1000MS
Memory Limit: 65536K

Total Submissions: 13849
Accepted: 4851

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).

【翻译】给出n个物品，每个物品有两个值a和b，选择n-k个元素，询问的最大值。

题解：

        ①题目要求求出Sigma式子的最大值，可以考虑单个式子ai/bi的最大值，然后将它们合起来。

        ②但是直接计算是不方便转化的,因为Sigma和单个式子还是有区别的。

        ③由于具有取值上的单调性，因此考虑二分,二分最大值x,那么则有：

                 ∑ai/∑bi>=x      移项得到：   ∑ai>=x*∑bi ——> ∑ai-x*∑bi>=0

        ④所以就二分啊，使得x不断变大,大到使得∑ai-x*∑bi几乎等于0，就是最有解了。

#include<stdio.h>
#include<algorithm>
#define go(i,a,b) for(int i=a;i<=b;i++)
#define ro(i,a,b) for(int i=a;i>=b;i--)
int n,k,a[4001],b[4001];
double T[4001],res,l,r,M;
bool check(double x)
{
    go(i,1,n)T[i]=a[i]-x*b[i];std::sort(T+1,T+1+n);res=0;
    go(i,k+1,n)res+=T[i];return res>=0;
}
int main()
{
    while(scanf("%d%d",&n,&k),n|k)
    {
    	go(i,1,n)scanf("%d",a+i);l=0;
    	go(i,1,n)scanf("%d",b+i);r=1;
    	while(r-l>1e-6)M=(l+r)/2,check(M)?l=M:r=M;printf("%.0f
",l*100);
    }
}//Paul_Gudeiran

 

希望你把我记住你流浪的孩子，无论在何时何地我都想念着你。————汪峰《我爱你中国》