我有一个程序，对于所有小于或等于输入值的整数，它可以找到可以表示为两个立方体之和的数字，两次，也就是Ramanujan的数字问题。
我已经用Java和Rust编写了这篇文章，但是，与Java相比，它在Rust中运行的速度慢了两倍多。
我能做些什么来使它表现得更好，或者改进它吗？
rust eclipse 代码：

use num_integer::Roots;
fn main() {
    let v = 984067;
    // let v = 87539319;
    for i in 1..=v {
        ramanujan(i)
    }
}
fn ramanujan(m: i32) {
    let maxcube = m.cbrt();
    let mut res1 = 0;
    let mut res2 = 0;
    let mut _res3 = 0;
    let mut _res4 = 0;
    for i in 1..=maxcube {
        for j in 1..=maxcube {
            if i * i * i + j * j * j == m {
                res1 = i;
                res2 = j;
                break;
            }
        }
    }
    for k in 1..=maxcube {
        for l in 1..=maxcube {
            if k == res1 || k == res2 || l == res1 || l == res2 {
                continue;
            }
            if k * k * k + l * l * l == m {
                _res3 = k;
                _res4 = l;
                break;
            }
        }
    }
    // if ((res1 * res1 * res1) + (res2 * res2 * res2) == m) && ((res3 * res3 * res3) + (res4 * res4 * res4) == m) {
    //     println!("{} is representable as the sums of two different sets of two cubes!\nThese values are {}, {}, and {}, {}.", m, res1, res2, res3, res4);
    // }
}

Java代码：

public class Ramun {
    public static void main(String[] args) {
        int v = 984067;
        // int v = 87539319;
        for (int i = 1; i <= v; i++) {
            ramanujan(i);
        }
    }

    public static void ramanujan(int m) {
        int maxcube = (int) Math.round(Math.cbrt(m));
        int res1 = 0, res2 = 0, res3 = 0, res4 = 0;
        for (int i = 1; i <= maxcube; i++) {
            for (int j = 1; j <= maxcube; j++) {
                if (((i * i * i) + (j * j * j)) == m) {
                    res1 = i;
                    res2 = j;
                    break;
                }
            }
        }
        for (int k = 1; k <= maxcube; k++) {
            for (int l = 1; l <= maxcube; l++) {
                if (k == res1 || k == res2 || l == res1 || l == res2)
                    continue;
                if (((k * k * k) + (l * l * l)) == m) {
                    res3 = k;
                    res4 = l;
                    break;
                }
            }
        }
        // if (((res1 * res1 * res1) + (res2 * res2 * res2) == m) && ((res3 * res3 * res3) + (res4 * res4 * res4) == m)) {
        //     System.out.printf("%d is representable as the sums of two different sets of two cubes!%nThese values are %d, %d, and %d, %d.%n", m, res1, res2, res3, res4);
        // }
    }
}

Time output for both programs