codefoces Zero Quantity Maximization (map + gcd)详解

D. Zero Quantity Maximization

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given two arrays aa and bb , each contains nn integers.

You want to create a new array cc as follows: choose some real (i.e. not necessarily integer) number dd , and then for every i∈[1,n]i∈[1,n] let ci:=d⋅ai+bici:=d⋅ai+bi .

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Your goal is to maximize the number of zeroes in array cc . What is the largest possible answer, if you choose dd optimally?

Input

The first line contains one integer nn (1≤n≤2⋅1051≤n≤2⋅105 ) — the number of elements in both arrays.

The second line contains nn integers a1a1 , a2a2 , ..., anan (−109≤ai≤109−109≤ai≤109 ).

The third line contains nn integers b1b1 , b2b2 , ..., bnbn (−109≤bi≤109−109≤bi≤109 ).

Output

Print one integer — the maximum number of zeroes in array cc, if you choose dd optimally.

题目的大概意思就是给你一串a[i], 一串b[i]， 让你找到一个数d，使c[i] = a[i] * d + b[i]为0的数量最多。

由于a[i] * d + b[i] = 0，所以d = -b[i] / a[i]. 若又两组a, b具有相同的d,则它们的最简分式相同。所以本题就变成了求最大公约数。

以下是代码

#include <iostream>
#include <algorithm>
#include <math.h>
#include <string.h>
#include <vector>
#include <stdio.h>
#include <map>
using namespace std;
#define LL long long
typedef pair<int, int> pii;
const int MAX = 2e5 + 50;

int gcd(int a, int b){
	return b == 0? a : gcd(b, a % b);
}
map<pii, int> mp;

int a[MAX];
int b[MAX];

int main(int argc, char const *argv[])
{
	int n;
	scanf("%d", &n);
	for(int i = 0; i < n; i++){
		scanf("%d", &a[i]);
	}

	for(int i = 0; i < n; i++){
		scanf("%d", &b[i]);
	}

	int cnt = 0;
	for(int i = 0; i < n; i++){
		if(a[i] != 0){
			if(a[i] < 0){// 因为是用map存，像-1，4 和1， -4这样的情况会被视为不相同，所以若a为负数，则变为正数，b变为相反数
				a[i] = -a[i];
				b[i] = -b[i];
			}
			int x = gcd(a[i], abs(b[i]));
			pii p;
			p.first = a[i] / x;
			p.second = b[i] / x;
			mp[p]++;
		} else{
			if(b[i] == 0){ // 注意 0， 0的情况，不管b去什么，结果都为0
				cnt++;
			}
		}
	}

	int ans = 0;
	for(auto x: mp){
		ans = max(ans, x.second); //取最大的情况
	}

	printf("%d\n", ans + cnt); // 加上cnt
	return 0;
}