数学方法，本题通过消元法转换为一元函数，但是编程不好实现，N个约束的转换
编程把每种材料作为最小的哪个，取所有情况最小值

import java.util.Scanner;

public class Main {
    public static void main(String[] args){
        Scanner scanner = new Scanner(System.in);
        int n = scanner.nextInt();
        double v = scanner.nextDouble();
        double[] a = new double[n];//比例
        double[] b = new double[n]; //上限
        double sum = 0;
        for(int i = 0; i < n; i++){
            a[i] = scanner.nextInt();
            sum += a[i];
        }
        for(int i = 0; i < n; i++){
            b[i] = scanner.nextInt();
            v = Math.min(v, b[i] * sum / a[i]);
        }
        System.out.println(String.format("%.4f", v));
    }
}