Bignum C++

#include <iostream>
#include <stdio.h>
#include <string>
#include <string.h>
#include <math.h>
#include <queue>
#include <memory.h>
using namespace std;

#define bigNum_size 220  //大数的段数，bigNum能表示的最大长度 = bigNum_size*log(mod) 
#define mod 100000   //每段存储数字的长度为0的个数 

char A[1010], B[1010];

struct bigNum
{
    int a[bigNum_size];
    int len;//存储大数的段数，初始化时长度都为1，表示的值默认为0 
    bigNum()
    {
        len = 1;
        memset(a, 0, sizeof(a));
    }
    bigNum(char *str)
    {
        int n = strlen(str);
        int t = 0;
        int m, i, k = 0, inf = 1;
        memset(a, 0, sizeof(a));
        for (i = n - 1; i >= 0; i--)
        {
            m = str[i] - '0';
            k += inf*m;
            inf *= 10;
            if (inf == mod)
            {
                inf = 1;
                a[t++] = k;
                k = 0;
            }
        }
        a[t++] = k;
        while (t > 1 && a[t-1] == 0) t--;
        len = t;
    }
    bigNum(int x)
    {
        memset(a, 0, sizeof(a));
        a[0] = x;
        int i = 1;
        while (a[i-1] >= mod)
        {
            a[i] = a[i-1]/mod;
            a[i-1] %= mod;
            i++;
        }
        len = i;
    }
};



bigNum operator +(bigNum x, bigNum y)
//重载运算符"+"，执行两个大数的加法并返回结果 
{
    int i, n;
    bigNum s;
    if (x.len < y.len)
    {
        bigNum temp = x;
        x = y;
        y = temp;
    }
    n = y.len;
    for (i = 0; i < n; i++)
    {
        x.a[i] += y.a[i];
        x.a[i+1] += x.a[i]/mod;
        x.a[i] %= mod;
    }
    while (x.a[n] > mod)
    {
        x.a[n+1] += x.a[n]/mod;
        n++;
    }
    if (x.a[x.len])x.len++;
    return x;
}



bigNum operator *(int m, bigNum x)
//重载运算符"*"，执行一个大数乘以一个int内的数字，并返回结果 
{
    int i,  k = 0;
    for (i = 0; i < x.len; i++)
    {
        x.a[i] = x.a[i] * m + k;
        k = x.a[i] / mod;
        x.a[i] %= mod;
    }
    while (k)
    {
        x.a[i] = k%mod;
        k /= mod;
        i++;
    }
    x.len = i;
    return x;
}

bool operator < (bigNum x, bigNum y)
//比较两个大数的大小，如果x<y返回true,否则返回false 
{
    if (x.len < y.len)
        return true;
    if (x.len > y.len)
        return false;
    int i;
    for (i = x.len - 1; i >= 0; i--)
        if (x.a[i] < y.a[i])
            return true;
        else if (x.a[i] > y.a[i])
            return false;
        return false;
}


bigNum operator - (bigNum x, bigNum y)
//计算两个大数的减法，其中保证x>y。 
{
    bigNum s;
    int i;
    int k = 0;
    for (i = 0; i < y.len; i++)
    {
        if (x.a[i] == -1)
        {
            x.a[i] = mod - 1;
            x.a[i+1]--;
        }
        if (x.a[i] >= y.a[i])
            s.a[i] = x.a[i] - y.a[i];
        else
        {
            s.a[i] = mod + x.a[i] - y.a[i];
            x.a[i+1]--;
        }
    }
    while (i < x.len)
    {
        if (x.a[i] == -1)
        {
            x.a[i] = mod - 1;
            x.a[i+1]--;
        }
        s.a[i] = x.a[i];
        i++;
    }
    s.len = i;
    while (s.a[s.len - 1] == 0)
    {
        s.len--;
    }
    if (s.len == 0)
        s.len = 1;
    return s;
}


bigNum sqrt_bignum(bigNum x)
//开根号，只能对大的正整数开方求得整数不分，
//如果要对小数开方或对正整数开方求小数点后面的数字，
//可以将这个数扩大10^(2*k)倍在进行开方。
// 在调用开发函数时， mod = 10 才能完成。所以效率不是很高。 
{
    bigNum s, md;
    bigNum p;
    bigNum temp;
    int n = x.len;
    int k;
    int i;
    if (n%2)
    {
        if (x.a[n-1] > 8)
        {
            s.a[0] = 3;
            md.a[0] = x.a[n-1] - 9;
        }
        else if (x.a[n-1] > 3)
        {
            s.a[0] = 2;
            md.a[0] = x.a[n-1] - 4;
        }
        else
        {
            s.a[0] = 1;
            md.a[0] = x.a[n-1] - 1;
        }
        n--;
    }
    
    for (i = n - 1; i > 0; i = i - 2)
    {
        p.a[1] = x.a[i];
        p.a[0] = x.a[i-1];
        p.len = 2;
        md = (100 * md + p);    
        for (k = 1; k < 10; k++)
        {
            p.a[0] = k;
            p.len = 1;
            temp = 20*s + p;
            temp = k*temp;
            if (md < temp)
                break;
        }
        k--;
        p.a[0] = k;
        p.len = 1;
        temp = 20*s + p;
        temp = k*temp;
        md = md - temp;
        s = 10*s + p;    
    }
    return s;
}


void output(bigNum x)
//输出大数的函数 
{
    int i;
    int n = x.len;
    printf("%d", x.a[n-1]);
    for (i = n - 2; i >= 0; i--)
        printf("%05d", x.a[i]);// %d05d与mod大小有关,即每段的数字个数。 
    printf("\n");
}

int main()
{
    return 0;
}