题面

Description

You are given two arrays $a$ and $b$, each contains $n$ integers.

You want to create a new array $c$ as follows: choose some real (i.e. not necessarily integer) number $d$, and then for every $i \in [1, n]$ let $c_i := d \cdot a_i + b_i$.

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

Input

The first line contains one integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the number of elements in both arrays.

The second line contains $n$ integers $a_1$, $a_2$, ..., $a_n$ ($-10^9 \le a_i \le 10^9$).

The third line contains $n$ integers $b_1$, $b_2$, ..., $b_n$ ($-10^9 \le b_i \le 10^9$).

Output

Print one integer — the maximum number of zeroes in array $c$, if you choose $d$ optimally.

Examples

Input

5
1 2 3 4 5
2 4 7 11 3

Output

2

Input

3
13 37 39
1 2 3

Output

2

Input

4
0 0 0 0
1 2 3 4

Output

0

Input

3
1 2 -1
-6 -12 6

Output

3

Note

In the first example, we may choose $d = -2$.

In the second example, we may choose $d = -\dfrac{1}{13}$.

In the third example, we cannot obtain any zero in array $c$, no matter which $d$ we choose.

In the fourth example, we may choose $d = 6$.

题目大意

给出两个长度为 $n$ $(1 \leq n \leq 2 \cdot 10^{5})$ 的数组 $a$ 和 $b$ ,求实数 $d$ 使得对所有 $i \in [1, n]$ 计算下式

$$ c[i] = d \cdot a[i] + b[i] $$

数组 $c$ 中 $0$ 的数目尽可能多。

题解

分析

令 $d \cdot a[i] + b[i] = 0$,则有 $d = -\dfrac{b[i]}{a[i]}$ 。但 $d$ 会爆 double 精度,此时用 pair 维护 <-b[i] / gcd, a[i] / gcd> 即可。但此时要注意处理 b[i]a[i] 的正负号关系(e.g. $\dfrac{1}{1} = \dfrac{-1}{-1}$ ,但在直接按此存储则会导致被认作为不同的值)。也可用 long double 维护 $d$ 。之后用 map 统计出现次数最多的 d 即可。

注意特判 $a[i] = 0$ 的情况。此时有 $c[i] = b[i]$ ,若 $b[i] = 0$ ,则 $c[i]$ 必为 $0$。

代码

pair 维护

#include <bits/stdc++.h>
using namespace std;

#define ABS(x) ((x) > 0 ? (x) : -(x))

const int MAXN = 2e5 + 10;

int a[MAXN], b[MAXN];

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);

    int n = 0;
    while (cin >> n) {
        for (int i = 1; i <= n; i++)
            cin >> a[i];
        for (int i = 1; i <= n; i++)
            cin >> b[i];

        int ans = 0, cnt0 = 0;

        map<pair<int, int>, int> m;
        for (int i = 1; i <= n; i++) {
            if (a[i] == 0) {
                if (b[i] == 0)
                    cnt0++;
            } else {
                pair<int, int> p = make_pair(-b[i], a[i]);
                if (p.first < 0 || (p.first == 0 && p.second < 0)) {
                    p.first *= -1;
                    p.second *= -1;
                }
                int d = __gcd(ABS(p.first), ABS(p.second));
                p.first /= d;
                p.second /= d;

                if (m.find(p) == m.end())
                    m.insert(make_pair(p, 1));
                else
                    m[p]++;

                ans = max(ans, m[p]);
            }
        }

        cout << ans + cnt0 << endl;
    }

    return 0;
}

long double维护

#include <bits/stdc++.h>
using namespace std;

typedef long long          ll;
typedef unsigned long long ull;
typedef long double        ld;

const int MAXN = 2e5 + 10;

int a[MAXN], b[MAXN];

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);

    int n = 0;
    while (cin >> n) {
        for (int i = 1; i <= n; i++)
            cin >> a[i];
        for (int i = 1; i <= n; i++)
            cin >> b[i];

        map<ld, int> m;
        int          ans = 0, cnt0 = 0;
        for (int i = 1; i <= n; i++) {
            if (a[i] == 0) {
                if (b[i] == 0)
                    cnt0++;
            } else {
                ld d = -(ld)b[i] / a[i];

                if (m.find(d) == m.end())
                    m.insert(make_pair(d, 1));
                else
                    m[d]++;

                ans = max(ans, m[d]);
            }
        }

        cout << ans + cnt0 << endl;
    }

    return 0;
}

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