Dropping tests
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 18430 | Accepted: 6376 |
Description
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 bifor 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).
参考：https://blog.csdn.net/hzoi_ztx/article/details/54898323
https://www.cnblogs.com/proverbs/archive/2013/01/09/2853725.html
设：xi∈{0,1}，表示第i个二元组是否留下，p为比率，P为p的最大值，即比例的最大值（注意区分大P和小p）
则：p=sigma(ai*xi)/sigma(bi*xi) 其中sigma(xi)=n-k
显然：对于所有可能的取得的 p 的值，p<=P
即：对于所有可能的xi的组合，sigma(ai*xi)/sigma(bi*xi)<=P
即：sigma(ai*xi)/sigma(bi*xi)的最大值等于P
即：sigma(ai*xi)-sigma(P*bi*xi)<=0，也就是sigma(ai*xi)-sigma(P*bi*xi)的最大值等于0
由于：sigma(ai*xi)-sigma(p*bi*xi)等价于sigma((ai-bi*p)*xi)，sigma((ai-bi*P)*xi)的最大值就等于0
同理：
设：我们枚举的p的最大值为m（对于所有的xi的组合，不一定存在m这个取值）
若m>P，sigma((ai-bi*P)*xi)的最大值小于0
若m<P，sigma((ai-bi*P)*xi)的最大值大于0
然后通过二分逼近的方法可以求出m=P~
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <queue>
#include <stack>
#include <set>
#include <map>
#define INF 0x3f3f3f3f
#define LL long long
using namespace std;
const int maxn=1005;
int n,k;
int a[maxn],b[maxn];
double l,r,d[maxn];
int main(){
while(~scanf("%d%d",&n,&k)&&(n+k)) {
double ans=0;
l=0,r=1;
for(int i=0;i<n;i++)
scanf("%d",&a[i]);
for(int i=0;i<n;i++)
scanf("%d",&b[i]);
double mid;
while(r-l>1e-8){
ans=0;
mid=(l+r)/2;
for(int i=0;i<n;i++)
d[i]=a[i]-mid*b[i];
sort(d,d+n);
for(int i=k;i<n;i++){
ans+=d[i];
}
if(ans>0) l=mid;
else r=mid;
}
printf("%.0lf\n",mid*100);
}
return 0;
}