POJ 2976:Dropping tests 01 分数规划

Dropping tests

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 15 0 25 1 6 4 2 1 2 7 9 5 6 7 9 0 0

Sample Output

83100

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).

POJ竟然用不了读入优化。。。努WA好多发

01分数规划裸题

#include<cmath>
#include<ctime>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<complex>
#include<iostream>
#include<algorithm>
#include<iomanip>
#include<vector>
#include<string>
#include<queue>
#include<set>
#include<map>
using namespace std;
typedef double db;
inline int read()
{
	int x=0,f=1;char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
	while(ch<='9'&&ch>='0'){x=(x<<1)+(x<<3)+ch-'0';ch=getchar();}
	return x*f;
}
const db eps=1e-6;
const int N=1010;
int n,m;
db d[N],a[N],b[N];
db work(db x)
{
	for(int i=1;i<=n;i++)d[i]=a[i]-x*b[i];
	sort(d+1,d+n+1);db sum=0.0;
	for(int i=m+1;i<=n;i++)sum+=d[i];
	return sum;
} 
int main()
{
	while(scanf("%d%d",&n,&m),n||m)
	{
		for(int i=1;i<=n;i++)scanf("%lf",&a[i]);
		for(int i=1;i<=n;i++)scanf("%lf",&b[i]);
		db l=0.0,r=1.0;
		while(fabs(r-l)>eps)
		{
			db mid=(l+r)/2;
			if(work(mid)>0)l=mid;
			else r=mid;
		}
		printf("%.0lf\n",l*100);
	}
	return 0;
}
/*
3 1
5 0 2
5 1 6
4 2
1 2 7 9
5 6 7 9
0 0

83
100

*/