大整数运算

Date：2019-07-11 19:42:01

算法实现

/*
Data: 2019-05-03 15:02:26
Problem:
*/

//存储
struct bign
{
    int d[M];
    int len;
    bign()
    {
        fill(d,d+M,0);
        len=0;
    }
};

//字符串->大数
bign Change(char str[])
{
    bign a;
    a.len = strlen(str);
    for(int i=0; i<a.len; i++)
        a.d[i] = str[a.len-i-1]-'0';    //字符串逆置存储
    return a;
}

//比较
int Compare(bign a, bign b)
{
    if(a.len > b.len)
        return 1;
    else if(a.len < b.len)
        return -1;
    else
    {
        for(int i=a.len-1; i>=0; i--)
            if(a.d[i] > b.d[i]) return 1;
            else if(a.d[i] < b.d[i])    return -1;
        return 0;
    }
}

//高精度加法
bign Add(bign a, bign b)
{   //其中之一为负数则采用减法，均为负数则相加再添符号
    bign c;
    int carry=0;    //进位
    for(int i=0; i<a.len||i<b.len; i++)
    {
        int temp = a.d[i]+b.d[i]+carry;
        c.d[c.len++] = temp%10;
        carry = temp/10;
    }
    if(carry != 0)
        c.d[c.len++] = carry;
    return c;
}

//A+B
#include<stdio.h>
#include<string.h>
const int M=1e3;
struct bign
{
    int d[M];
    int len;
    bign()
    {
        memset(d,0,sizeof(d));
        len=0;
    }
};

//小端存储
bign Change(char str[])
{
    bign a;
    a.len = strlen(str);
    for(int i=0; i<a.len; i++)
        a.d[i] = str[a.len-i-1]-'0';
    return a;
}

bign Add(bign a, bign b)
{
    bign c;
    int carry=0;
    for(int i=0; i<a.len||i<b.len; i++)
    {
        int temp = a.d[i]+b.d[i]+carry;
        c.d[c.len++] = temp%10;
        carry = temp/10;
    }
    if(carry != 0)
        c.d[len++] = carry;
    return c;
}

void Print(bign a)
{
    for(int i=a.len-1; i>=0; i--)
        printf("%d", a.d[i]);
}

int main()
{
    char str1[M], str2[M];
    scanf("%s%s", str1,str2);
    bign a = Change(str1);
    bign b = Change(str2);
    Print(add(a,b));
    return 0;
}

//高精度减法
bign Sub(bign a, bign b)
{   //若a<b，交换a,b，添负号
    bign c;
    for(int i=0; i<a.len||i<b.len; i++)
    {
        if(a.d[i] < b.d[i])
        {
            a.d[i+1]--;
            a.d[i]+=10;
        }
        c.d[c.len++] = a.d[i]-b.d[i];
    }
}

//高精度与低精度的乘法    bign * int
bign Multi(bign a, int b)
{   //负数取绝对值再相乘，高精度乘高精度，则外面套一层bign b的循环即可
    bign c;
    int carry=0;
    for(int i=0; i<a.len; i++)
    {
        int temp = a.d[i]*b + carry;
        c.d[c.len++] = temp%10;
        carry = temp/10;
    }
    while(carry != 0)
    {
        c.d[c.len++] = carry%10;
        carry /= 10;
    }
    return c;
}

//高精度与低精度的除法
bign Divide(bign a, int b, int &r)  //余数为r,初值=0
{
    bign c;
    c.len = a.len;  //被除数与商的各位一一对应
    for(int i=a.len-1; i>=0; i--)
    {
        r = r*10 + a.d[i];
        if(r < b)   c.d[i]=0;   //不够除
        else
        {
            c.d[i] = r/b;
            r %= b;
        }
    }
    while(c.len>1 && c.d[c.len-1]==0)
        c.len--;        //去除高位的0，若全是0，则保留一位0
    return c;
}