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 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
解题思路：
贪不了，学习了一下新姿势orz—直接二分答案。
代码：
#include<iostream>
#include<stdio.h>
#include<algorithm>
#define maxn 1003
using namespace std;
const double eps=1e-7;
inline int read(){
int x=0,f=0;char ch=getchar();
while(ch>'9'||ch<'0')f|=ch=='-',ch=getchar();
while(ch>='0'&&ch<='9')x=(x<<3)+(x<<1)+(ch^48),ch=getchar();
return f?-x:x;
}
int n,k,a[maxn],b[maxn];
double d[maxn];
bool cmp(double a,double b){return a>b;}
int main(){
while(1){
n=read(),k=read();
if(!n&&!k)break;
for(int i=0;i<n;++i)a[i]=read();
for(int i=0;i<n;++i)b[i]=read();
double l=0.0,r=1.0,mid;
while (r-l>eps){
mid=(l+r)/2;
for(int i=0;i<n;++i) d[i]=a[i]-mid*b[i];
sort(d,d+n,cmp);
double sum=0.0;
for(int i=0;i<n-k;++i) sum+=d[i];
if(sum>0) l=mid;
else r=mid;
}
printf("%.0f\n",mid*100);
}
return 0;
}