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，并在计算出后将其乘积写入输出文件中。
数字是“大数字”，它们不能存储在变量int或long中，因此我想将它们的数字存储在数组中。
我写了下面的代码，该代码似乎可以处理一些数字，但是在插入例如
时失败55545252575453525348514954525256515056485755545157575748525349575251565256515554545052494854495550525752505756 * 50485456554954525356575657575048545450535257534951545354504957957454525485485565565525525155515656575655535754
应该是
2804227435732034912849259183651914853320425721942955023578511952364152651715628400479161029284275995992691222252865738869930419556645027062168262691246978645407150215825953157986304093759319409280056139599351010978148800024
我知道
28042274357320349128382480725408037422083045098207428013562897291318294193822950671357266858499225582466777983203943354354852
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();
}
}
我知道还有许多其他算法可以做同样的事情，并且比我的算法要好得多，但是我想了解为什么这并不总是有效。预先感谢。
答案 0 :(得分：1)
您的错误发生在您要从其中单元格中位数超过一位的单元格进行最终结转的部分：
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]最多包含两位数字（不大于99）。该假设是不正确的。我尝试使用您的程序将99 999 999 999（十一位数）与其自身相乘，此时res[12]包含100。因此，(res[i] / 10) % 10的计算结果为0，因此您没有像往常那样携带任何东西到res[11]或res[10]。
我想我应该自己找一个好的解决方法。
编辑：为了便于阅读，这是您自己的注释的格式和缩进格式的修正：
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
}
}