java - 大数乘法算法并不总是有效

Question

我正在尝试创建一个函数，该函数从输入文件中读取两个正自然数，并在计算后将其乘积写入输出文件。
这些数字是“大数字”，它们不能存储在变量中，例如int我long想将它们的数字存储在数组中。

我编写了以下代码，该代码似乎适用于某些数字，但例如在插入时失败

55545252575453525348514954525256515056485755545157575748525349575251565256515554545052494854495550525752505756 * 50485456554954525356575657575048555450535257534951545354504957495454525448485751565254525155515656575655535754

那应该是

2804227435732034912849259183651914853320425721942955023578511952364152651715628400479161029284275992691222252865738869930419556645027062168262691246978645407150215825953157986304093759319409280056139599351010978148800024

我明白了

2804227435732034912838248072540803742208304509820742801356289729131829419382295067135726685849922558246677798320394335486985213210683628834929357923746322174827983613731945865091982648208298268945129498351010978148800024

这是代码：

public static void multiplication(String inFileName, String outFileName) {
        int[] a, b, res; //a is the first number, b is the second number, res is the final result
        int a_i, b_i; //indexes to navigate inside the numbers

        a = readBigNumber(inFileName, 0); //This works
        if(a == null) return;
        System.out.print("*** a: "); printIntArray(a); System.out.println("");  //DEBUG

        b = readBigNumber(inFileName, 1); //This works
        if(b == null) return;
        System.out.print("*** b: "); printIntArray(b); System.out.println("");  //DEBUG

        res = new int[a.length + b.length + 1]; //The result cannot be longer than this
        Arrays.fill(res, 0); //Fill the array with zeros

        //if the length of a < length of b, swap
        int[] tempArr;
        if (a.length < b.length) {
            tempArr = new int[a.length];
            System.arraycopy(a, 0, tempArr, 0, a.length);
            a = new int[b.length];
            System.arraycopy(b, 0, a, 0, b.length);
            b = new int[tempArr.length];
            System.arraycopy(tempArr, 0, b, 0, tempArr.length);
        }

        //this algorithm works like the "manual" column multiplication
        int temp;
        int res_i = 0; //index to navigate in the res array
        int carry = 0;
        for(b_i = b.length-1; b_i >= 0; b_i--) {
            for(a_i = a.length-1; a_i >= 0; a_i--) {
                temp = a[a_i] * b[b_i] + carry; //save the product in a temp variable
                res_i = res.length - 1 - (a.length - 1 - a_i) - (b.length - 1 - b_i); //calculate the index in the res array
                //I need to have just one digit [0-9]
                if(temp > 9) { //If temp has more than one digit, take the right-one and put the left-one in the carry
                    res[res_i] += temp % 10; //right-digit
                    carry = (temp / 10) % 10; //left-digit
                } else {
                    res[res_i] += temp;
                    carry = 0;
                }
            }

            //when I exit the a-loop, if the carry is not 0, I have to put it in the result before continuing
            if(carry > 0) {
                res[res_i - 1] += carry;
                carry = 0;
            }
        }

        //Once completed, each array cell could have more than one digit.
        //Check it right to left and if so, keep the right-digit e sum the left-digit in the left cell
        for(int i = res.length-1; i >= 0; i--) {
            if(res[i] > 9) {
                res[i - 1] += (res[i] / 10) % 10; //left-digit
                res[i] = res[i] % 10; //right-digit
            }
        }

        //Write the final result
        String res_str = "";
        int start_i = 0;
        while(start_i < res.length && res[start_i] == 0) start_i++; //Ignore initial zeros
        if(start_i == res.length) {
            writeBigNumber(outFileName, "0"); //I checked the whole array, the product is 0
        } else {
            for(int i = start_i; i < res.length; i++) {
                res_str += String.valueOf(res[i]);
            }
            writeBigNumber(outFileName, res_str);
        }

        System.out.println("*** res: " + res_str);  //DEBUG
    }

这些是一些“辅助功能”：

//This works
private static int[] readBigNumber(String inFileName, int lineIndex) {
        File inputFile = new File(inFileName);
        Scanner scan = null;
        int i = 0;
        int[] num;
        char[] rawInput;

        try {
            scan = new Scanner(inputFile);
            while(scan.hasNextLine()) {
                rawInput = scan.nextLine().toCharArray();
                if(i == lineIndex) {
                    num = new int[rawInput.length];
                    for(int j = 0; j < num.length; j++) {
                        num[j] = rawInput[j] - '0';
                    }
                    scan.close();
                    return num;
                }
                i++;
            }
        } catch (Exception e) {
            System.out.println("An error occurred while reading the numbers from file: ");
            e.printStackTrace();
        } finally {
            if(scan != null)
                scan.close();
        }

        return null;
    }

    private static void writeBigNumber(String outFileName, String num) {
        try {
            File outputFile = new File(outFileName);
            FileWriter writer = new FileWriter(outputFile, false);
            writer.write(num);
            writer.close();
        } catch (Exception e) {
            System.out.println("An error occurred while writing the result: ");
            e.printStackTrace();
        }
    }

我知道有许多其他算法可以做同样的事情并且比我的要好得多，但我想了解为什么这并不总是有效。提前致谢。

标签： javamultiplicationbignum

Answer 1

您的错误在于您正在从具有多个数字的单元格中进行最终结转：

        if(res[i] > 9) {
            res[i - 1] += (res[i] / 10) % 10; //left-digit
            res[i] = res[i] % 10; //right-digit
        }

在这里，您假设res[i]最多有两位数（不大于 99）。假设不成立。我尝试使用您的程序将 99 999 999 999（11 位数字）与自身相乘，res[12]此时包含 100。所以(res[i] / 10) % 10评估为 0，你没有携带任何东西res[11]或res[10]你应该携带任何东西。

我想我应该让自己找到一个好的解决方法。

编辑：为了可读性，这是您自己的评论，格式化和缩进的修复：

    for (int i = res.length - 1; i >= 0; i--) {
        if (res[i] > 9) {
            if (res[i] > 99) {
                res[i - 1] += ((res[i] / 100) % 10) * 10; // first digit (*10)
                res[i - 1] += (res[i] / 10) % 10; // second digit
            } else {
                res[i - 1] += (res[i] / 10) % 10; // left-digit
            }
            res[i] = res[i] % 10; // right-digit
        }
    }