Array Division - What is the best way to divide two numbers stored in an array?

问题

I have two arrays (dividend, divisor):

dividend[] = {1,2,0,9,8,7,5,6,6};
divisor[] = {9,8};

I need the result (dividend/divisor) as:

quotient[] = {1,2,3,4,5,6,7};

I did this using array subtraction:

subtract divisor from dividend until dividend becomes 0 or less than divisor, each time incrementing quotient by 1,

but it takes a huge time. Is there a better way to do this?

回答1:

Is there a better way to do this?

You can use long division.

回答2:

Do long division.

Have a temporary storage of size equal to the divisor plus one, and initialized to zero:

accumulator[] = {0,0,0};

Now run a loop:

Shift each digit of the quotient one space to the left.
Shift each digit of the accumulator one space to the right.
Take the next digit of the dividend, starting from the most-significant end, and store it to the least-significant place of the accumulator.
Figure out accumulator / divisor and set the least-significant place of the quotient to the result. Set the accumulator to the remainder.

Used to use this same algorithm a lot in assembly language for CPUs what didn't have division instructions.

回答3:

Other than that, have you considered using BigInt class (or the equivalent thing in your language) which will already does this for you?

回答4:

You can use long division http://en.wikipedia.org/wiki/Long_division

回答5:

You can use Long division algorithm or the more general Polynomial Long Division.

回答6:

Why not convert them to integers and then use regular division?

in pseudocode:

int dividend_number
foreach i in dividend
    dividend_number *= 10
    dividend_number += i

int divisor_number
foreach i in divisor
    divisor_number *= 10
    divisor_number += i

int answer = dividend_number / divisor_number;

回答7:

There you go! A is the divident. B is the divisor. C is the integer quotinent R is the rest. Each "huge" number is a vector retaining a big number. In huge[0] we retain the number of digits the number has and thren the number is memorized backwards. Let's say we had the number 1234, then the corespoding vector would be:

v[0]=4; //number of digits
v[1]=4;
v[2]=3;
v[3]=2;
v[4]=1;

....

SHL(H,1) does: H=H*10;
SGN(A,B) Compares the A and B numbers
SUBSTRACT(A,B) does: A=A-B;
DIVIDHUGE: makes the division using the mentioned procedures...

void Shl(Huge H, int Count)
/* H <- H*10ACount */
{ 
  memmove(&H[Count+1],&H[1],sizeof(int)*H[0]);
  memset(&H[1],0,sizeof(int)*Count);
   H[0]+=Count;
}
    int Sgn(Huge H1, Huge H2) {
      while (H1[0] && !H1[H1[0]]) H1[0]--;
      while (H2[0] && !H2[H2[0]]) H2[0]--;

      if (H1[0] < H2[0]) {
        return -1;
      } else if (H1[0] > H2[0]) {
        return +1;
      }

      for (int i = H1[0]; i > 0; --i) {
        if (H1[i] < H2[i]) {
          return -1;
        } else if (H1[i] > H2[i]) {
          return +1;
        }
      }
      return 0;
    }

        void Subtract(Huge A, Huge B)
        /* A <- A-B */
        { int i, T=0;

          for (i=B[0]+1;i<=A[0];) B[i++]=0;
          for (i=1;i<=A[0];i++)
            A[i]+= (T=(A[i]-=B[i]+T)<0) ? 10 : 0;
          while (!A[A[0]]) A[0]--;
        }


            void DivideHuge(Huge A, Huge B, Huge C, Huge R)
            /* A/B = C rest R */
            { int i;

              R[0]=0;C[0]=A[0];
              for (i=A[0];i;i--)
                { Shl(R,1);R[1]=A[i];
                  C[i]=0;
                  while (Sgn(B,R)!=1)
                    { C[i]++;
                      Subtract(R,B);
                    }
                }
              while (!C[C[0]] && C[0]>1) C[0]--;
            }

来源：https://stackoverflow.com/questions/3322129/array-division-what-is-the-best-way-to-divide-two-numbers-stored-in-an-array

标签

c++

algorithm

arrays

digits