大数运算

1、加法

//blog.csdn.net/simplestbest/article/details/6788298

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

#define MAXSIZE 1000

void bigadd(const char *a, const char *b, char *pret)
{
    int ia = strlen(a)-1; //a的个位索引
    int ib = strlen(b)-1; //b的个位索引
    int carry = 0; //进位
    int sum = 0; // 每位的和
    int ret_len = ia > ib ? (ia+1) : (ib+1);
    // malloc时考虑进位和'\0'
    char *result = (char *)malloc( (ret_len + 2) * sizeof(char) );
    if( !result ) /* perror(); */ exit(-1);
    ret_len = 0;

    while( ia >= 0 || ib >= 0)
    {
        sum = carry;
        if(ia >= 0)
        {
            sum += a[ia--] - '0';
        }
        if(ib >= 0)
        {
            sum += b[ib--] - '0';
        }
        carry = sum/10;
        sum = sum%10;
        result[ret_len++] = sum + '0';
    }
    if(carry > 0)
        result[ret_len++] = carry + '0';
    //result[ret_len] = '\0'; Not useful.
    while(ret_len > 0)
    {
        *pret++ = result[--ret_len];
    }
    *pret = '\0';

    free(result);
}

int main(int argc, char *argv[])
{
    if (argc != 3)
    {
        printf("Usage: %s a b\n", argv[0]);
        return -1;
    }

    char result[MAXSIZE] = {0};
    bigadd(argv[1], argv[2], result);
    printf(" %s\n + %s\n", argv[1], argv[2]);
    printf("--------------------------------\n %s", result);
    return 0;
}

3、乘法

  未测，写的有点乱。

void bigMultiple(char *a, char *b, char *pret)
{
    int xArr[MAXSIZE], yArr[MAXSIZE], result[MAXSIZE*2];
    int len_a = strlen(a);
    int len_b = strlen(b);
    int i,j;

    for(i = 0; i < MAXSIZE; ++i)
    {
        xArr[i] = 0;
        yArr[i] = 0;
    }
    for(i = 0; i < MAXSIZE*2; ++i)
    {
        result[i] = 0;
    }

    // atoi and reversed into Arr
    for(i = 0; i < len_a; ++i)
    {
        //a[0]是高位，作为xArr的最后一个数
        xArr[len_a - 1 - i] = a[i] - '0';
    }
    for(j = 0; j < len_b; ++j)
    {
        yArr[len_b - 1 - j] = b[j] - '0';
    }

    // calculate
    for(i = 0; i < len_a; ++i)
    {
        for(j = 0; j < len_b; ++j)
        {
            result[i+j] += a[i]*b[j];
        }
    }
    for(i = 0; i < (len_a + len_b + 1); ++i)
    {
        if(result[i] > 0)
        {
            result[i+1] += result[i]/10; //进位
            result[i] = result[i]%10;
        }
    }
    // return to pret, use prev i.
    for(; i >= 0 ; --i)
    {
        if(result[i] != 0)
            break;
    }

    for(j = 0; j < i; ++j)
        *(pre+j) = result[i-j];
    *(pre+j) = '\0';
}

2、减法

 

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