容易發現答案只與w1和w2的比值有關，而與具體數值無關

那麼先特殊算一下w1=0和w2等於0的情況，然後就直接假設w1=1

然後的話對於每個物品可能有4種情況：永遠比1號優，永遠比1號劣，永遠和1號相等，當w2<某值時比1號優，等於時相等，否則比一號劣，或者當w2>某值時比1號劣，等於時相等，否則比1號優

對於每個手機算一下他的情況和w2的分界點，然後按分界點排序之後掃一遍即可

比賽的時候有個地方x和y打反一直wa，然後就棄療了

#include<iostream>
#include<cstdlib>
#include<cstdio>
#include<cstring>
#include<ctime>
#include<cmath>
#include<algorithm>
#include<iomanip>
#include<queue>
#include<map>
#include<bitset>
#include<stack>
#include<vector>
#include<set>
using namespace std;
#define MAXN 100010
#define MAXM 1010
#define INF 1000000000
#define MOD 1000000007
#define ll long long
#define eps 1e-8
struct data{
	double v;
	int c;
	friend bool operator <(data x,data y){
		return x.v<y.v;
	}
};
int n;
int x[MAXN],y[MAXN];
int low=1,high=INF;
data t[MAXN];
int tot;
int main(){
	int i;
	scanf("%d",&n);
	for(i=1;i<=n;i++){
		scanf("%d%d",&x[i],&y[i]);
	}
	int tlowx=1,thighx=1,tlowy=1,thighy=1;
	for(i=2;i<=n;i++){
		if(x[i]>x[1]){
			thighx++;
		}
		if(x[i]>=x[1]){
			tlowx++;
		}
		if(y[i]>y[1]){
			thighy++;
		}
		if(y[i]>=y[1]){
			tlowy++;
		}
	}
	low=max(low,max(tlowx,tlowy));
	high=min(high,min(thighx,thighy));
	int tlow=1,thigh=1;
	for(i=2;i<=n;i++){
		if(y[i]==y[1]){
			if(x[i]>=x[1]){
				tlow++;
			}
			if(x[i]>x[1]){
				thigh++;
			}
			continue ;
		}
		if(y[i]<y[1]){
			t[++tot].v=1000.*(x[1]-x[i])/(y[i]-y[1]);
			t[tot].c=-1;
			tlow++;
			thigh++;
		}
		if(y[i]>y[1]){
			t[++tot].v=1000.*(x[1]-x[i])/(y[i]-y[1]);
			t[tot].c=1;
		}
		if(t[tot].v<=0){
			tlow+=t[tot].c;
			thigh+=t[tot].c;
			tot--;
		}
	}
	if(tot){
		sort(t+1,t+tot+1);
		for(i=1;i<=tot;){
			int wzh=i;
			int tc0=0,tc1=0;
			while(wzh<=tot&&fabs(t[wzh].v-t[i].v)<eps){
				if(t[wzh].c==1){
					tc1++;
				}else{
					tc0++;
				}
				wzh++;
			}
			low=max(low,tlow+tc1);
			high=min(high,thigh-tc0);
			tlow+=tc1-tc0;
			thigh+=tc1-tc0;
			i=wzh;
		}
	}
	printf("%d %d\n",high,low);
	return 0;
}

/*
2
3 2
1 2

*/