所以这是一个家庭作业，我一直在做约10个小时。我想知道一些提示，看看我哪里出错了。所以我的任务是基本上为大整数做一个计算器，我的教授提供了骨架。他要求我们写的函数是减法，除法，乘法，取幂（x ^ n）和一个小函数。我很确定我的LessThan和multiply函数正常工作，但也许不行，反正我的工作在下面，我会在之后描述我的问题：
public class Calc {
//
// main
//
// Parses the command-line arguments as an integer
// operation, packages the operands as digit arrays,
// calls the appropriate calculator function, and
// then outputs the result.
//
public static void main(String[] args) {
if (args.length == 3) {
int[] input1 = construct(args[0]);
int[] input2 = construct(args[2]);
if (args[1].equals("plus")) {
int[] result = plus(input1,input2);
putln(result);
} else if (args[1].equals("minus")) {
int[] result = minus(input1,input2);
putln(result);
} else if (args[1].equals("times")) {
int[] result = times(input1,input2);
putln(result);
} else if (args[1].equals("div")) {
int[] result = div(input1,input2);
putln(result);
} else if (args[1].equals("power")) {
int[] result = power(input1,input2);
putln(result);
} else if (args[1].equals("less")) {
boolean result = lessThan(input1,input2);
TextIO.putln(result);
} else if (args[1].equals("equal")) {
boolean result = equal(input1,input2);
TextIO.putln(result);
} else {
TextIO.putln("Invalid binary operation. Try using +, -, +, /, ^, <, or =.");
}
} else if (args.length == 2) {
int[] input = construct(args[0]);
// it must be a unary operation
if (args[1].equals("!")) {
int[] result = factorial(input);
TextIO.putln(result);
} else {
TextIO.putln("Expected factorial operation!");
}
} else {
TextIO.putln("Invalid input. Try either a binary operation or factorial.");
}
}
//
// plus
//
// Computes the sum of two digit arrays,
// represented as a digit array.
//
public static int[] plus(int[] a, int[] b) {
// Determine the result's maximum length.
//
int alen, blen, len;
alen = length(a);
blen = length(b);
if (alen > blen) {
len = alen + 1;
} else {
len = blen + 1;
}
// Allocate an array of digits with that length.
//
int[] c = new int[len];
// Compute the sum of the digits, working from
// least to most significant.
//
int carry = 0;
for (int i = 0; i < len; i = i+1) {
int sum = carry;
if (i < alen) {
sum = sum + a[i];
}
if (i < blen) {
sum = sum + b[i];
}
c[i] = sum % 10;
carry = sum / 10;
}
// Trim off the leading 0s.
//
return trim(c);
}
// INCOMPLETE
//
// minus
//
// Computes the difference of two digit arrays,
// represented as a digit array.
//
// You can assume that the first operand is
// larger than the second one.
//
public static int[] minus(int[] a, int[] b) {
if(equal(a,b))
{
int [] zero = new int [1];
zero[0] = 0;
return zero;
}
else{
int c = 0;
int lb = length(b);
while(lb > c)
{
b[c] = (b[c]) * (-1);
c++;
}
return plus(a,b);
}
}
// INCOMPLETE
//
// times
//
// Computes the product of two digit arrays,
// represented as a digit array.
//
// HINT: use repeated addition. You might need
// to write lessThan first.
//
// BONUS: perform this using the schoolbook
// method.
//
public static int[] times(int[] a, int[] b) {
int len;
if(length(a) > length(b))
len = length(a) + 1;
else
len = length(b) + 1;
int[] c = new int[len];
int[]counterArray;
if(lessThan(a,b))
{
counterArray = new int[length(b)];
counterArray[length(b)-1] = 1;
}
else
{
counterArray = new int[length(a)];
counterArray[length(a)-1] = 1;
}
int[] counterTwo = new int[1];
counterTwo[0] = 1;
while(lessThan(counterArray,a))
{
c = plus(c,b);
counterArray = plus(counterArray, counterTwo);
}
return plus(c,b);
}
// INCOMPLETE
//
// div
//
// Computes the quotient of two digit arrays,
// represented as a digit array.
//
public static int[] div(int[] a, int[] b) {
int[] counter = zero();
boolean value = true;
if(lessThan(a,b))
return zero();
else {
while(value)
{
a = minus(a,b);
counter = plus(counter,one());
if(lessThan(b,a))
value = false;
}
return counter;
}
/**boolean checker = false;
int[] c = new int[length(a)];
for(int i = 0; i < length(a); i++)
{
c[i] = 0;
}
int counter = 0;
while (!checker)
{
if(equal(a,c))
{
checker = true;
}
else if(lessThan(a,c))
{
checker = true;
}
else if (lessThan(c,a))
{
c = plus(c,b);
counter++;
}
}
return construct("" + (counter));*/
}
//
// factorial
//
// Computes the factorial of a digit array,
// as a digit array.
//
public static int[] factorial(int[] a) {
int[] count = one();
int[] product = one();
while (lessThan(count,a) || equal(count,a)) {
product = times(product,count);
count = plus(count,one());
}
return product;
}
// INCOMPLETE
//
// power
//
// Computes the p-th power of a digit array x,
// where p is also given as a digit array. The
// result is calculated as a digit array.
//
public static int[] power(int[] x, int[] p) {
int pl = length(p);
int[] checker = new int[1];
checker[0] = 0;
int[] adder = new int[1];
adder[0] = 0;
if ((p[pl - 1] % 2) == 0)
{
while(equal(checker,p) == false)
{
adder = plus(square(x),adder);
p = minus(p,construct("" + 2));
}
return adder;
}
else
{
int[] checkertwo = new int[1];
checkertwo[0] = 1;
while(equal(checkertwo,p) == false)
{
adder = plus(square(x), adder);
p = minus(p,construct("" + 2));
}
return times(adder,x);
}
}
public static int[] square(int[] x)
{
return times(x,x);
}
//
// equal
//
// Returns whether or not two digit arrays
// represent the same number.
//
public static boolean equal(int[] a, int[] b) {
int[] ta = trim(a);
int[] tb = trim(b);
if (ta.length != tb.length) {
return false;
} else {
for (int i = a.length-1; i >= 0; i = i-1) {
if (a[i] != b[i]) {
return false;
}
}
return true;
}
}
// INCOMPLETE
//
// lessThan
//
// Returns whether or not one digit array
// represents a number less than another.
//
public static boolean lessThan(int[] a, int[] b) {
boolean value = false;
a = minus(a,construct(""+0));
b = minus(b,construct(""+0));
if(length(b) > length(a))
{
value = true;
}
else if(length(a) == length(b))
{
boolean c = false;
int counter = 0;
while(counter < length(a))
{
c = false;
if(a[counter] > b[counter])
{
c = true;
value = false;
}
if(!c)
value = true;
counter++;
}
}
else
value = false;
return value;
}
// * * * * * * * * * * * * * * * * * * * * * * * *
//
// These are helper functions that can be used in
// the above code.
//
//
// construct
//
// Converts a string that consists of decimal digits
// into an array of those digits.
//
public static int[] construct(String digits) {
int len;
int[] a;
len = digits.length();
a = new int[len];
// Copy each digit of the string into
// the digit array, but as integers
// rather than characters.
for (int i = len-1; i >= 0; i = i-1) {
// get the i-th digit
char c = digits.charAt(i);
// check if it is a valid digit
if (c >= '0' && c <= '9') {
a[len-1-i] = (int)(c-'0');
} else {
// treat nondigits as 0 digits
a[len-1-i] = 0;
}
}
// trim off leading 0s
return trim(a);
}
//
// zero
//
// returns the digits representing 0
//
public static int[] zero() {
return construct("0");
}
//
// one
//
// returns the digits representing 0
//
public static int[] one() {
return construct("1");
}
//
// trim
//
// Takes an array of integer digits representing an integer,
// one that may have leading 0 digits, and constructs an
// array with no leading digits.
//
public static int[] trim(int[] a) {
int len = length(a);
int[] b;
// build an array of that length
b = new int[len];
// set the digits
for (int i=0; i<len; i++) {
b[i] = a[i];
}
return b;
}
//
// put
//
// Outputs an array of digits, from most- to least-
// significant, excluding any leading 0s.
//
public static void put(int[] a) {
int len = length(a);
for (int i = len-1; i >= 0; i = i-1) {
TextIO.put(a[i]);
}
}
//
// putln
//
// Outputs an array of digits, from most- to least-
// significant, excluding any leading 0s, followed by
// a carriage return.
//
public static void putln(int[] a) {
put(a);
TextIO.putln();
}
//
// length
//
// Counts the number of digits in a digit array
// excluding leading 0s.
//
public static int length(int[] a) {
int len = a.length;
// figure out the number's real length
while (len > 0 && a[len-1] == 0) {
len = len - 1;
}
if (len == 0) {
return 1;
} else {
return len;
}
}
}
我的减号似乎不起作用，出于某种原因，我不知道出了什么问题。我觉得x-y = x +（ - y）是我认为我正确做的事情。也许我的教授所写的加功能对此不起作用？我试过想一想。而且我认为这个减函数对于我的除法和幂函数是不可或缺的。对不起，我的笔记不清楚，但是有人可以看一遍，并告诉我最重要的是我在哪里搞砸负功能，因为我认为一旦我得到正确的工作，我就能做到其余部分。
答案 0 :(得分：0)
此代码未经过测试，但应该可以使用：
public static int[] minus(int[] a, int[] b) {
int len = length(a);
int blen = length(b);
int[] c = new int[len];
int carry = 0;
for (int i = 0; i < len; i = i+1) {
int sum = a[i] - carry;
if (i < blen) sum -= b[i];
if (sum < 0) {
sum += 10;
carry = 1;
} else {
carry = 0;
}
c[i] = sum;
}
}
答案 1 :(得分：0)
在plus方法的末尾添加以下代码（在 for 循环中）。
carry = sum / 10; // existing
if (c[i] < 0) {
c[i] = c[i] + 10; // Create a positive complement of base 10.
carry = carry - 1; // Borrowed. (or carry = -1)
}
如果没有此修复，当第二个操作数位数较大时，它将返回一个负值，该值在结果中显示为 - digit 。例如，12 + (-8)将打印为1-6