两个整数数组的成对乘法。 如果它们的长度不同，则较短的需要缠绕

Question

I have two integer arrays.

A[] = {3,4,5,6}
B[] = {1,2}

How can we multiply these two arrays in order to get a product of 3456*12 using long multiplication?

Answer 1

Very crude example in Python.

a = [3,4,5,6]
b = [1,2]
result = []                        
for i in range(0, len(a)):
     result.append(a[i % len(a)]*b[i % len(b)])

print result

Answer 2

In Java

 public static int mul(int[] a, int []b) {
      int res=0, i, j, mcol, mrow;
      for (mcol=1, i=a.length-1 ; i>=0 ; i--, mcol*=10) {
            for (mrow=1, j=b.length-1 ; j>=0 ; j--, mrow*=10) {
                 res += a[i] * b[j] * mcol * mrow;
            }
      }
      return res;
 }

Principle is the same as school multiplication

3 4 5 6
x   1 2

=          2x6 + 2x5x10 + 2x4x100 +2x3x1000
   + 10 x (1x6 + 1x5x10 + 1x4x100 +1x3x1000)

Starting from 6 , multiply by 2 , then 2 x 5 in 2 ^nd position, so you have to *10 it, and add it to the result, and so on...

For the next digit, 1 , do the same operation, but multiply each time this result again by 10 since 1 is the next digit in the lower row.

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You can also consider the multiplication from the left side of the arrays.

 public static int mul(int[] a, int []b) {
      int res=0, resrow, i, j;
      for (i=0 ; i<a.length ; i++) {
            for (resrow=j=0 ; j<b.length ; j++) {
                 resrow = resrow * 10 + a[i] * b[j];
            }
            res = res * 10 + resrow;
      }
      return res;
 }

the function is slightly different

  public static int mul(int[] a, int []b) { int res=0, resrow, i, j; for (i=0 ; i<a.length ; i++) { for (resrow=j=0 ; j<b.length ; j++) { resrow = resrow * 10 + a[i] * b[j]; } res = res * 10 + resrow; } return res; } 

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Call like this

  public static void main(String[] args) { int a[] = {3,4,5,6}; int b[] = {1,2}; int r = mul(a,b); out.println(r); }