大数相乘算法

[c]代码库

/**************************************
	大数相乘算法

    算法复杂度为：O(longhta*longthb）
    longtha为乘数的位数
    longhtb为被乘数的位数
***************************************/

#include <stdio.h>415
#include <string.h>
#include <conio.h>
#define LEN 1000
void mult(char [], char [], char []);
main()
{
    char op1[LEN], op2[LEN], op3[LEN * 2 - 1];
    scanf("%s%s", op1, op2);
    mult(op1, op2, op3);
    printf("%s\n", op3);
    getch();
    return 0;
}
void reverse(char a[])
{
    int longth = strlen(a);
    int i;
    for(i = 0; i < longth / 2; i++)
    {
        char t;
        t = a[i];
        a[i] = a[longth - i - 1];
        a[longth - i - 1] = t;
    }
}
void mult(char op1[LEN], char op2[LEN], char ans[LEN * 2 - 1])
{
    char top1[LEN];
    char top2[LEN];
    strcpy(top1, op1);
    strcpy(top2, op2);
    reverse(top1);
    reverse(top2);
    int k;
    int top1s = strlen(top1);
    int top2s = strlen(top2);
    for(k = 0; k < top1s + top2s; k++)
    {
        ans[k] = '0';
    }
    int i, j;
    int jw, ys;
    int longth;
    for(j = 0; j < top2s; j++)
    {
        jw = 0;
        for(i = 0; i < top1s; i++)
        {
            ys = ((top1[i] - '0') * (top2[j] - '0') + jw + ans[i + j] - '0') % 10;
            jw = ((top1[i] - '0') * (top2[j] - '0') + jw + ans[i + j] - '0') / 10;
            ans[i + j] = ys + '0';
        }
        if(jw > 0)
        {
            ans[i + j] = jw + '0';
        }
    }
    longth = i + j - 1;
    if(jw > 0)
        ans[longth++] = jw + '0';
    ans[longth] = '\0';
    reverse(ans);
}

[代码运行效果截图]