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【C】两个大整数的乘法

作者: / 发布于2014/5/22/ 447

这只是两个大整数的乘法，加法，减法应不成问题，你可参照下面的程序要是除法的话，可以用数值分析里的牛顿逼近法


	/*这只是两个大整数的乘法，加法，减法应不成问题，你可参照下面的程序
	要是除法的话，可以用数值分析里的牛顿逼近法*/

	/* * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * *
	* Question: *
	* How to calculate two integers each one has forty bit. *
	* Algorithm: *
	* Declare two arrays suppose to be array A and B to storage *
	* the two integers just as a intacter singly. *
	* Get the first number from array B,then use it to multiply *
	* the number of array A singly. Declare another array C to *
	* storage the value when multiplying. *
	* Thinking about the carrying bits. *
	* Thinking: *
	* Can you use the Linear_List to do this? *
	* If they are two floating numbers,so how to do it? *
	* * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * */

	#include <stdio.h>
	#include <assert.h>
	#include <stdlib.h>
	#include <conio.h>

	//To calculate two 40 bits integers multiply,change 10 to 40
	#define THE_ARR_A_SIZE 10 #define ARR_B_SIZE THE_ARR_A_SIZE
	#define ARR_C_SIZE (THE_ARR_A_SIZE + ARR_B_SIZE)

	void main(void);
	void Multiply( short *, short *, short *, short, short);
	void Carrying( short *c, short);

	void main(void)
	{
	 short A[THE_ARR_A_SIZE] = {9, 9, 9, 9, 9, 9, 9, 9, 9, 9}; // 10
	 short B[ARR_B_SIZE] = {9, 9, 9, 9, 9, 9, 9, 9, 9, 9};
	 short C[ARR_C_SIZE];
	 short i = 0;

	 for ( NULL; i < ARR_C_SIZE; i++)
	 {
	 *(C+i) = 0;
	 }

	 Multiply(A, B, C, THE_ARR_A_SIZE, ARR_B_SIZE);
	 Carrying(C, ARR_C_SIZE);

	 printf("\\nThe Result is: ");

	 i = 0;
	 if (C[0] == 0) ++i; //To delete the primacy zero
	 for ( NULL; i < ARR_C_SIZE; i++)
	 {
	 printf("%d", C[i]);
	 }
	 getch();
	 printf("\\n");
	}

	/*
	* NOTICE:
	* The Array C is not initialed,you should do that. Here
	* the array is supposed to be filled zero.
	* Algorithm:
	*
	*/

	void Multiply( short A[], short B[], short C[], short A_size, short B_size)
	{
	 short i = 0, j = 0, k = 1; // Assigned k to 1 to leave C[0]
	 //be null to storage the carrying.

	 for (j = 0; j<B_size; j++)
	 {
	 for (i = 0; i<A_size; i++)
	 {
	 C[k] += B[j] * A[i];
	 k++;
	 }
	 k = 2;
	 k = k + j;
	 }
	}

	// To solve the carrying .

	void Carrying(short C[],short C_size)
	{
	 short i, tens_place = 0, units_order = 0;
	 short temp = 0, end_arr;

	 end_arr = C_size-1;
	 for (i = end_arr; i >= 0; i--)
	 {
	 temp = C[i];
	 if (C[i] > 9)
	 {
	 tens_place = temp / 10;
	 units_order = temp % 10;
	 C[i-1] += tens_place;
	 C[i] = units_order;
	 }
	 }
	}

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