乘法運算可以分拆為兩步：第一步，是將乘數與被乘數逐位相乘；第二步，將逐位相乘得到的結果，對應相加起來。

public class BigInt {  
    public static void main(String[] args) {  
        Scanner sc = new Scanner(System.in);  
        String str1 = sc.next();  
        String str2 = sc.next();  
        int[] num1 = new int[str1.length()];  
        int
 
[] num2 = new int[str2.length()];  
        //把字串轉換成int陣列
        for (int i = 0; i < str1.length(); i++) {  
            num1[str1.length() - 1 - i] = str1.charAt(i) - '0';  
        }  
        for (int i = 0; i < str2.length(); i++) {  
            num2[str2.length() - 1 - i] = str2.charAt(i) - '0'
 
;  
        }  

        int[] result = multiply(num1, num2);  
        // 因為前面做了倒序操作，所以最高位在最後
        for (int i = result.length - 1; i >= 0; i--) {  
            if(i==result.length - 1)  
                if(result[i]==0)  
                    continue;  
            System.out.print(result[i]);  
        }  

        sc.close();  
    }  

    public
 
 static int[] multiply(int[] num1, int[] num2) {  
        int lengthOfNum1 = num1.length;  
        int lengthOfNum2 = num2.length;  
        //如果num1和num2的長度分別為n1，n2，那麼它們相乘的結果位數最大為n1+n2
        int[] result = new int[lengthOfNum1 + lengthOfNum2];  
        //num[i]*num2[j]的結果存在result[i+j]上，最後再處理進位制問題  
        for (int i = 0; i < lengthOfNum1; i++) {  
            for (int j = 0; j < lengthOfNum2; j++) {  
                result[i + j] += num1[i] * num2[j];  
            }  
        }  
        // 處理進位制問題  
        for (int i = 0; i < lengthOfNum1 + lengthOfNum2 - 1; i++) {  
            if (result[i] >= 10) {  
                result[i + 1] += result[i] / 10;  
                result[i] = result[i] % 10;  
            }  
        }  
        return result;  
    }  
}