所以我不能让变量可以整除,我需要能够做到这一点,否则我不知道如何完成构建我想要构建的锁。
它使用20个输入的数字,然后将它们排列成代数的方程2 /微积分系统,然后通过删除“e”来解决“s”,“a”,“f”和“e”从等式中取代。
我非常感谢帮助,我对这些想法持开放态度,因为我有25个这样的构建,这只是第一个的1/3。
简而言之,我如何划分变量?
import java.util.Scanner;
public class Lock
{
public static void main(String[] args) {
Scanner user_input = new Scanner (System.in);
String num_a;
System.out.print("Enter the first number: ");
num_a = user_input.next();
String num_b;
System.out.print("Enter the second number: ");
num_b = user_input.next();
String num_c;
System.out.print("Enter the third number: ");
num_c = user_input.next();
String num_d;
System.out.print("Enter the fourth number: ");
num_d = user_input.next();
String num_e;
System.out.print("Enter the fifth number: ");
num_e = user_input.next();
String num_f;
System.out.print("Enter the sixth number: ");
num_f = user_input.next();
String num_g;
System.out.print("Enter the seventh number: ");
num_g = user_input.next();
String num_h;
System.out.print("Enter the eigth number: ");
num_h = user_input.next();
String num_i;
System.out.print("Enter the ninth number: ");
num_i = user_input.next();
String num_j;
System.out.print("Enter the tenth number: ");
num_j = user_input.next();
String num_k;
System.out.print("Enter the eleventh number: ");
num_k = user_input.next();
String num_l;
System.out.print("Enter the twetlth number: ");
num_l = user_input.next();
String num_m;
System.out.print("Enter the thirteenth number: ");
num_m = user_input.next();
String num_n;
System.out.print("Enter the fourteenth number: ");
num_n = user_input.next();
String num_o;
System.out.print("Enter the fifteenth number: ");
num_o = user_input.next();
String num_p;
System.out.print("Enter the sixteenth number: ");
num_p = user_input.next();
String num_q;
System.out.print("Enter the seventeenth number: ");
num_q = user_input.next();
String num_r;
System.out.print("Enter the eighteenth number: ");
num_r = user_input.next();
String num_s;
System.out.print("Enter the nineteenth number: ");
num_s = user_input.next();
String num_t;
System.out.print("Enter the twentieth number: ");
num_t = user_input.next();
System.out.println(num_a + "s + " + num_b + "a + " + num_c + "f + " + num_d + "e = " + num_e);
System.out.println(num_f + "s + " + num_g + "a + " + num_h + "f + " + num_i + "e = " + num_j);
System.out.println(num_k + "s + " + num_l + "a + " + num_m + "f + " + num_n + "e = " + num_o);
System.out.println(num_p + "s + " + num_q + "a + " + num_r + "f + " + num_s + "e = " + num_t);
System.out.println(num_a + "s + " + num_b + "a + " + num_c + "f + " + num_d + "[(" + num_t + " " + num_p + "s + " + num_q + "a " + num_r + "f) / " + num_s + "] =" + num_e);
System.out.println(num_f + "s + " + num_g + "a + " + num_h + "f + " + num_i + "[(" + num_t + " " + num_p + "s + " + num_q + "a " + num_r + "f) / " + num_s + "] =" + num_j);
System.out.println(num_k + "s + " + num_l + "a + " + num_m + "f + " + num_n + "[(" + num_t + " " + num_p + "s + " + num_q + "a " + num_r + "f) / " + num_s + "] =" + num_o);
// THIS creates the fourth equation items/order to be substituted into the other first three equations.
int t = num_t;
int s = num_s;
int num_ts = (t / s);
num_ts =
num_ps = (num_p / num_s);
num_qs = (num_q / num_s);
num_rs = (num_r / num_s);
// THIS is the Fourth equation being substituted into the First Equation
num_dts = (num_d * num_ts);
num_dps = (num_d * num_ps);
num_dqs = (num_d * num_qs);
num_drs = (num_d * num_rs);
// THIS is the Fourth equation being substituted into the Second Equation
num_its = (num_i * num_ts);
num_ips = (num_i * num_ps);
num_iqs = (num_i * num_qs);
num_irs = (num_i * num_rs);
// THIS is the fourth equation being substituted into the Third Equation
num_nts = (num_n * num_ts);
num_nps = (num_n * num_ps);
num_nqs = (num_n * num_qs);
num_nrs = (num_n * num_rs);
System.out.println(num_a + "s + " + num_b + "a + " + num_c + "f + " + num_dts + " " + num_dps + "s + " + num_dqs + "a " + num_drs + "f = " + num_e);
System.out.println(num_f + "s + " + num_g + "a + " + num_h + "f + " + num_its + " " + num_ips + "s + " + num_iqs + "a " + num_irs + "f = " + num_j);
System.out.println(num_k + "s + " + num_l + "a + " + num_m + "f + " + num_nts + " " + num_nps + "s + " + num_nqs + "a " + num_nrs + "f = " + num_o);
}
}
答案 0 :(得分:1)
您无法添加,减去,除以或乘以字符串变量。您必须将变量设置为整数才能执行此操作。此外,您可以使用数组来保存变量,因为它们有很多变量。
答案 1 :(得分:0)
String
,Integer
,Float
的类型不同。例如,您无法在/
上应用*
或String
等运算符。 +
很特殊,因为它有String
的定义,这意味着连接。
由于您需要对用户输入执行某些操作,因此您可以直接将其读取为int
:
System.out.print("Enter the first number: ");
int num_a = user_input.nextInt();
System.out.print("Enter the second number: ");
int num_b = user_input.nextInt();
然后你可以做
int num_ab = a / b;
请注意,如果a < b
,则num_ab
将为0
,因为这是int
eger。您可能想要执行类似
float num_ab = (float)a / b;
现在,这段代码非常繁琐。如果您接受处理索引而不是变量的字母,则可以在循环中初始化它们,例如
Scanner in = new Scanner(System.in);
int[] numbers = new int[20];
int index = 0;
while (index < numbers.length) {
System.out.println("Enter the "+(index+1)+"th number");
int n = in.nextInt();
numbers[index] = n;
index++;
}
System.out.println(Arrays.toString(numbers));
使用数字数组
// arrays start at 0
int num_ab = numbers[0] / numbers[1];
如果您希望能够通过名称访问变量,则可以定义常量
static final int a = 0;
static final int b = 1;
static final int c = 2;
//...
int num_ab = numbers[a] / numbers[b];
但在您的情况下,将初始变量和计算的变量存储在某些可以检索它们以进行进一步计算的地方可能很方便:
// the store for all the variables and their value
static Map<String, Integer> vars = new HashMap<>();
// the function to read in the store
static Integer var(String name) {
return vars.get(name);
}
商店通过循环初始化:
Scanner in = new Scanner(System.in);
// The 20 variables...
String alpha = "abcdefghijklmnopqrst";
for (char c : alpha.toCharArray()) {
String varName = String.valueOf(c);
System.out.println("Enter the value for "+ varName);
int n = in.nextInt();
vars.put(varName, n);
}
System.out.println(vars.toString());
int num_ab = var("a")/var("b");
// Store ab for further computation
vars.put("ab", num_ab);
System.out.println("ab is " + var("ab");