题解 | #幻兽防御战#

#include <iostream>
#include <vector>
#include <algorithm>

using namespace std;

// 定义怪兽结构体
struct Monster {
    long long d, s;
};

// 按到达时间排序：d1/s1 < d2/s2 <=> d1 * s2 < d2 * s1
bool compareMonsters(const Monster& a, const Monster& b) {
    return a.d * b.s < b.d * a.s;
}

int main() {
    int n;
    cin >> n;
    vector<long long> d(n), s(n);
    for (int i = 0; i < n; ++i) cin >> d[i];
    for (int i = 0; i < n; ++i) cin >> s[i];

    vector<Monster> monsters(n);
    for (int i = 0; i < n; ++i) {
        monsters[i] = {d[i], s[i]};
    }

    // 按照到达时间升序排序
    sort(monsters.begin(), monsters.end(), compareMonsters);

    int ans = 0;
    for (int k = 0; k < n; ++k) {
        // 判断第 k 只怪兽是否在第 k 次射击前或射击瞬间到达
        // 比较 D_k / S_k <= k
        if (monsters[k].d <= (long long)k * monsters[k].s) {
            break;
        }
        ans++;
    }

    cout << ans << endl;

    return 0;
}

import java.util.*;

public class Main {
    // 怪兽类
    static class Monster {
        long d, s;
        Monster(long d, long s) {
            this.d = d;
            this.s = s;
        }
    }

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        long[] d = new long[n];
        long[] s = new long[n];
        for (int i = 0; i < n; i++) d[i] = sc.nextLong();
        for (int i = 0; i < n; i++) s[i] = sc.nextLong();

        List<Monster> monsters = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            monsters.add(new Monster(d[i], s[i]));
        }

        // 排序规则：d1/s1 < d2/s2 即 d1*s2 < d2*s1
        Collections.sort(monsters, (a, b) -> {
            long val1 = a.d * b.s;
            long val2 = b.d * a.s;
            return Long.compare(val1, val2);
        });

        int ans = 0;
        for (int k = 0; k < n; k++) {
            Monster m = monsters.get(k);
            // 检查 D_k <= k * S_k
            if (m.d <= (long) k * m.s) {
                break;
            }
            ans++;
        }
        System.out.println(ans);
    }
}

def solve():
    # 读取怪兽数量
    n = int(input())
    # 读取初始距离
    d_list = list(map(int, input().split()))
    # 读取移动速度
    s_list = list(map(int, input().split()))

    monsters = []
    for i in range(n):
        monsters.append((d_list[i], s_list[i]))

    # 按照到达时间 d/s 升序排序
    # Python 自动处理大整数，可以直接使用 d/s 排序或用 d * other_s 比较
    monsters.sort(key=lambda x: x[0] / x[1])

    ans = 0
    for k in range(n):
        d, s = monsters[k]
        # 判断第 k 只怪兽是否在第 k 分钟准备射击时或之前到达
        if d <= k * s:
            break
        ans += 1
    
    print(ans)

solve()