Zero Quantity Maximization

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 DEBUGfreopen("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 DEBUGprintf("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 longusing 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;
}

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