【C语言】超大数乘法运算

昨天做排列组合的时候遇到A(a,b)这个问题，需要计算A（20,20）超大，计算机32位的，最大数只能是2^32，这让我很悲伤！

于是乎就自己研究了如何进行超大数的计算！

/*********************************************************************************
 *      Copyright:  (C) 2013 Chen ZhenWei<[e-mail]>
 *                  All rights reserved.
 *
 *       Filename:  bignum.c
 *    Description:  This file
 *
 *        Version:  1.0.0(12/12/2013~)
 *         Author:  Chen ZhenWei <[e-mail]>
 *      ChangeLog:  1, Release initial version on "12/12/2013 08:21:23 AM"
 *
 ********************************************************************************/

#include <stdio.h>
#include <string.h>

int save_num(const char *str,unsigned short *num)
{
   int i=0;
   int length = strlen(str);

   for(i=0; i<length/2; i++)
   {
    //   printf("%c == %c\n",str[length-1-i*2],str[length-2-i*2]);
       num[i] = (str[length-1-i*2]-'0')+10*(str[length-2-i*2]-'0');
   }
   if(length%2 != 0)
   {
       num[i] = str[0] - '0';
       return i+1;
   }
   return i;
}

void main()
{
    unsigned short numa[100];
    unsigned short numb[100];
    unsigned short result[200];
    unsigned short carry;
    int i=0,j=0,lena,lenb,res_len;
    char strnum[64];
    char flag;

    scanf("%s",strnum);
    lena = save_num(strnum,numa);
    //scanf("%c",&flag);
    scanf("%s",strnum);
    lenb = save_num(strnum,numb);
    memset(result,0x0,200*2);

    for(i=0; i<lena; i++)
        for(j=0; j<lenb; j++)
        {
           carry = numa[j] * numb[i];
           result[j+i] += carry%100;
           result[j+i+1] += carry/100;
        }
    res_len = j+i+1; 

    for(i=0; i<res_len; i++)
    {
        if(result[i]>=100)
        {
            result[i+1] += result[i]/100;
            result[i] = result[i]%100;
        }
    }
    for(i=res_len; i>0; i--)
    {
        printf("%02d",result[i-1]);
    }

    printf("\n");

}

设计思路：

操作结果：