poj2976 Dropping tests 01分数规划

\[\sum a_i - x\times \sum b_i \]

等同于

\[\sum(a_i-x \times b_i) \]

#include <algorithm>
#include <iostream>
#include <cstdio>
using namespace std;
int n, k, uu;
double a[1005], b[1005], c[1005], l, r;
bool chk(double x){
	for(int i=1; i<=n; i++)	c[i] = a[i] - x * b[i];
	sort(c+1, c+1+n);
	double sum=0.0;
	for(int i=n; i>=k+1; i--)	sum += c[i];
	return sum>0;
}
int main(){
	while(scanf("%d %d", &n, &k)!=EOF && (n || k)){
		for(int i=1; i<=n; i++){
			scanf("%d", &uu);
			a[i] = uu;
		}
		for(int i=1; i<=n; i++){
			scanf("%d", &uu);
			b[i] = uu;
		}
		l = 0.0, r = 1.0;
		while(r-l>1e-6){
			double mid=(l+r)/2.0;
			if(chk(mid))	l = mid;
			else	r = mid;
		}
		printf("%.0f\n", 100.0*l);
	}
	return 0;
}