https://codeforces.com/contest/1133/problem/D

题解：就是求斜率相同的最大个数

特判a[i]  b[i]为0的时候三种情况，其中a[i]  b[i]同时为0时的情况可以加到答案里

C++版本一

long double 有精度问题

/*
*@Author:   STZG
*@Language: C++
*/
#include <bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<string>
#include<vector>
#include<bitset>
#include<queue>
#include<deque>
#include<stack>
#include<cmath>
#include<list>
#include<map>
#include<set>
//#define DEBUG
#define RI register int
using namespace std;
typedef long long ll;
//typedef __int128 lll;
const int N=200000+10;
const int M=100000+10;
const int MOD=1e9+7;
const double PI = acos(-1.0);
const double EXP = 1E-8;
const int INF = 0x3f3f3f3f;
int t,n,m,k,q;
int ans,cnt,flag,temp,sum;
int a[N];
int b[N];
long double c[N];
char str;
struct node{};
int main()
{
#ifdef DEBUG
	freopen("input.in", "r", stdin);
	//freopen("output.out", "w", stdout);
#endif
    //ios::sync_with_stdio(false);
    //cin.tie(0);
    //cout.tie(0);
    //scanf("%d",&t);
    //while(t--){
    scanf("%d",&n);
    for(int i=1;i<=n;i++){
        scanf("%d",&a[i]);
    }
    for(int i=1;i<=n;i++){
        scanf("%d",&b[i]);
    }
    for(int i=1;i<=n;i++){
        if(a[i]==0){
            if(b[i]!=0)
                c[i]=-INF;
            else{
                cnt++;
                c[i]=INF;
            }
            continue;
        }
        c[i]=(long double)b[i]/(long double)a[i];
    }
    sort(c+1,c+n+1);
    for(int i=1;i<=n;i++){
        if(c[i]==-INF||c[i]==INF)
            continue;
        ans=max(ans,(int)(upper_bound(c+1,c+n+1,c[i])-lower_bound(c+1,c+n+1,c[i])));
    }
    cout<<ans+cnt<<endl;
    //}

#ifdef DEBUG
	printf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC);
#endif
    //cout << "Hello world!" << endl;
    return 0;
}

C++版本二

分数解法

#include <bits/stdc++.h>
#define ll long long

using namespace std;

int aa[200005],ab[200005];
struct node
{
    ll a,b;
}rr[200005];
bool cmp(node a1, node a2)
{
    if(a1.a == a2.a){
        return a1.b < a2.b;
    }
    return a1.a < a2.a;
}

int main()
{
    int n;
    scanf("%d",&n);
    int cnt = 0, d0 = 0, a0 = 0, b0 = 0;
    for(int i = 0; i < n; ++i)
    {
        scanf("%d",&aa[i]);
    }
    for(int i = 0; i < n; ++i)
    {
        scanf("%d",&ab[i]);
        if(aa[i] == 0)
        {
            ++a0;//�κ�d��������
            if(ab[i] == 0)
            {
                ++d0;//�κ�d������
                ++b0;//d����0������
            }
        }else if(ab[i] == 0){
            ++b0;
        }else{
            ll g = __gcd(aa[i],ab[i]);
            rr[cnt].a = (long long)aa[i]/g;
            rr[cnt].b = (long long)ab[i]/g;
            if(rr[cnt].a < 0){
                rr[cnt].a *= -1;
                rr[cnt].b *= -1;
            }
            cnt++;
        }
    }
    sort(rr,rr+cnt,cmp);
    int ans = 0,cc = 1;
    if(cnt > 0){
        ans = 1;
    }
    for(int i = 1; i < cnt; ++i){
        if(rr[i-1].a * rr[i].b == rr[i-1].b * rr[i].a){
            cc++;
            ans = max(ans,cc);
        }else{
            cc = 1;
        }
    }

    printf("%d\n",max(b0,d0+ans));
    return 0;
}