JAVA中的砖块问题

给出了一个用Java做的练习，但不知道如何做：

我们想制作一排X英寸长的砖，我们有一些1英寸长的小砖供用户输入，还有一些5英寸长的大砖供用户输入

我已经确定了目标的长度，然后是存储为int的每个砖块的数量，并取给定砖块数量的总长度，即smallbrick*1和bigbrick*确定是否可以使用输入的砖块数量使目标X英寸。5虽然我知道用文字做什么，我不知道如何将其转换为代码

代码如下：

import java.util.Scanner; 
public class BrickMan {

    public static void main(String[] args) {
        Scanner kb = new Scanner(System.in); 
        System.out.println("How Large is the Goal (In Inches)? "); 
        int goal = kb.nextInt(); 
        System.out.println("How Many Big Bricks Do You Have? "); 
        int bgbrk = kb.nextInt(); 
        System.out.println("How Many Small Bricks Do You Have? "); 
        int smlbrk = kb.nextInt(); 
        byte smlbrklen = 1; 
        byte bgbrklen = 5; 
        int ttlbgbrk = bgbrklen * bgbrk; 
        int ttlsmlbrk = smlbrklen * smlbrk; 

谢谢你

小砖 大砖块 X

---

Find sb听起来你应该先尝试一些代码。我们喜欢简单地看到一些代码，X-minX%5，大*5，你可以，resBig=minX%5，大；resSmall=X-5*resBig；堆栈溢出不存在，无法为您完成作业。仅供参考：

X == sb * 1 + lb * 5;

int lb = 0;
int sb = 0;
int builtLength = 0;
while (built has room for a large brick) {
    add a large brick
}
while (built has room for a small brick) {
    add a small brick
}

Integer division:    17 / 5 = 3
Integer modulus:     17 % 5 = 2
                     3 * 5 + 2 = 17

int totalLength=kb.nextInt()
int bigBricks = kb.nextInt();
int smallBricks = kb.nextInt();
int lengthOfBricks = bigBricks*5+smallBricks;
if (lengthOfBricks>=totalLength)
    System.out.println("It's long enough");
else
    System.out.println("It's not long enough");