poj2976

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题解

  • 裸题,注意精度
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#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<cstring>
#include<algorithm>
#define rep(i,x,y) for(register int i=x;i<=y;++i)
#define repd(i,x,y) for(register int i=x;i>=y;--i)
#define ll long long
using namespace std;
const int N=1e3+7;
const ll lnf=0x3f3f3f3;
const double esp=1e-3;
double a[N],b[N],c[N];
int n,m;
double check(double x){
	double ans=0;
	rep(i,1,n)c[i]=1ll*a[i]-1ll*x*b[i];
	sort(c+1,c+n+1);
	repd(i,n,m+1){
		ans+=c[i];
		if(ans<0)return -1;
	}
	return ans;
}
double find(){
	double l=0,r=1.0*lnf;
	while(l+esp<r){
		double mid=(l+r)/2;
		if(check(mid)>=0)l=mid;
		else r=mid;
	}
	return l;
}
int main(){
	while(~scanf("%d%d",&n,&m)&&(n||m)){
		rep(i,1,n){scanf("%lf",&a[i]);a[i]*=100;}
		rep(i,1,n)scanf("%lf",&b[i]);
		double ans=find();
		printf("%.0f\n",ans);
	}
	return 0;
}
  • Dinkelbach算法
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#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<cstring>
#include<algorithm>
#define rep(i,x,y) for(register int i=x;i<=y;++i)
#define repd(i,x,y) for(register int i=x;i>=y;--i)
#define ll long long
using namespace std;
const int N=1e3+7;
const ll lnf=0x3f3f3f3;
const double esp=1e-3;
struct node{
	double a,b,c;
	inline double f(double x){c=a-b*x;}
}a[N];
int n,m;
inline int cmp(node x,node y){return x.c<y.c;}
double solve(double x){
	rep(i,1,n)a[i].f(x);
	sort(a+1,a+n+1,cmp);
	double aa=0,bb=0;
	repd(i,n,m+1){aa+=a[i].a;bb+=a[i].b;}
	return aa/bb;
}
double find(){
	double l=0,r=solve(0);
	while(l+esp<r){
		l=r;r=solve(r);
	}
	return l;
}
int main(){
	while(~scanf("%d%d",&n,&m)&&(n||m)){
		rep(i,1,n){scanf("%lf",&a[i].a);a[i].a*=100;}
		rep(i,1,n)scanf("%lf",&a[i].b);
		double ans=find();
		printf("%.0f\n",ans);
	}
	return 0;
}