c++大整數類的幾種實現方法與解析

在oj上做題時，相信大家遇到過很多要求大整數的題目，這時候，我們就需要用到所謂的高精度算法，即用數組來儲存整數，模擬四則運算以及其他常見的運算，下面就讓我們來分析一下幾種實現大整數的方法

一.用vector來儲存大整數

<span style="font-size:24px;color:#FF0000;"><strong><span style="font-size:14px;color:#000000;">#include<iostream>
#include<vector>
#include<stack>
#include<cstring>
#include<string>
#include<cstdio>
using namespace std;
class BigInteger  
{
public:
	static const int base=100000000;
	static const int width=8;
	vector<int>s;
	BigInteger(long long num=0){*this=num;}    //構造函數
	BigInteger operator=(long long num)
	{
		s.clear();
		do{
			s.push_back(num%base);
			num=num/base;
		}while(num>0);
		return *this;
	}
	BigInteger operator=(const string &str)      //重載=號
	{
		s.clear();
		int x,len=(str.length()-1)/width+1;
		for(int i=0;i<len;i++){
			int end=str.length()-i*width;
			int start=max(0,end-width);
			sscanf(str.substr(start,end-start).c_str(),"%d",&x);    //格式符%d是讀入十進制整數
			                                                    //string.c_str是Borland封裝的String類中的一個函數，它返回當前字符串的首字符地址
			s.push_back(x);
		}
		return *this;
	}
	friend ostream & operator<<(ostream &out,const BigInteger& x)   //重載輸出號
	{
	out<<x.s.back();
	for(int i=x.s.size()-2;i>=0;i--){
		char buf[20];
		sprintf(buf,"%08d",x.s[i]);
		for(int j=0;j<int(strlen(buf));j++)
			out<<buf[j];
	}
	return out;
	}
	friend istream & operator>>(istream &in,BigInteger& x)   //重載輸入號
	{
	string s;
	if(!(in>>s)) return in;
	x=s;
	return in;
	}
	BigInteger operator+(const BigInteger& b)const   //重載加號
	{
		BigInteger c;
		c.s.clear();
		for(int i=0,g=0;;i++){
			if(g==0&&i>=s.size()&&i>=b.s.size()) break;
			int x=g;
			if(i<s.size()) x+=s[i];
			if(i<b.s.size()) x+=b.s[i];
			c.s.push_back(x%base);
			g=x/base;
		}
		return c;
	}
	BigInteger operator-(const BigInteger& b)   //重載減號,默認前面大於後面
	{
		BigInteger c;
		c.s.clear();
		if(*this>b){
			int i,g;
		for(i=0,g=0;;i++){
			if(g==0&&i>=b.s.size()) break;
			int x=g;
			if(s[i]<b.s[i]){
				s[i+1]-=1;
				s[i]=s[i]+base;
			}
			if(i<s.size()) x+=s[i];
			if(i<b.s.size()) x-=b.s[i];
			c.s.push_back(x%base);
			g=x/base;
		}
		int x=0;
		for(;i<s.size();i++){
			x+=s[i];
			c.s.push_back(x%base);
			x=x/base;
		}
		}
		return c;
	}
	bool operator<(const BigInteger& b)const   //重載小於號
	{
		if(s.size()!=b.s.size()) return s.size()<b.s.size();
		for(int i=s.size()-1;i>=0;i--)
			if(s[i]!=b.s[i]) return s[i]<b.s[i];
		return false;
	}
	bool operator>(const BigInteger& b)const   //重載大於號
	{
		return b<*this;
	}
	bool operator<=(const BigInteger& b)const
	{
		return !(b<*this);
	}
	bool operator>=(const BigInteger& b)const
	{
		return !(*this<b);
	}
	bool operator==(const BigInteger& b)const  //重載等於號
	{
		return !(b<*this)&&!(*this<b);
	}
	BigInteger operator+=(const BigInteger& b)
	{
		*this=(*this+b);
		return *this;
	}
	BigInteger operator-=(const BigInteger& b)
	{
		*this=(*this-b);
		return *this;
	}
};</span></strong></span>

這段代碼是我參照算法競賽入門上面的，下面我們來分析一下具體實現過程：

       首先定義了base=100000000，width=8，來實現大整數每8位用int的方式保存進vector內，然後通過重載賦值運算符，輸入字符串的形式實現將字符串轉換爲vector容器值。

       然後是加法的重載實現，通過每取八位出來進行+運算，若結果大於8位則進一位，並將這一位保留到下一次的運算中。

       而減法的重載實現，則是通過每取八位出來進行-運算，若結果小於0則從s[i+1]，也就是vector容器內後一個值中取出10000000來進行計算，而s[i+1]則減一來實現借位。

二.通過數組的方式儲存大整數

<span style="font-size:24px;color:#FF0000;"><strong><span style="color:#000000;"><span style="font-size:14px;">#include<string>
#include<iostream>
#include<iosfwd>
#include<cmath>
#include<cstring>
#include<stdlib.h>
#include<stdio.h>
#include<cstring>
#define MAX_L 2005 
using namespace std;

class bign
{
public:
    int len, s[MAX_L];//數的長度，記錄數組
//構造函數
    bign();
    bign(const char*);
    bign(int);
    bool sign;//符號 1正數 0負數
    string toStr() const;//轉化爲字符串，主要是便於輸出
    friend istream& operator>>(istream &,bign &);//重載輸入流
    friend ostream& operator<<(ostream &,bign &);//重載輸出流
//重載複製
    bign operator=(const char*);
    bign operator=(int);
    bign operator=(const string);
//重載各種比較
    bool operator>(const bign &) const;
    bool operator>=(const bign &) const;
    bool operator<(const bign &) const;
    bool operator<=(const bign &) const;
    bool operator==(const bign &) const;
    bool operator!=(const bign &) const;
//重載四則運算
    bign operator+(const bign &) const;
    bign operator++();
    bign operator++(int);
    bign operator+=(const bign&);
    bign operator-(const bign &) const;
    bign operator--();
    bign operator--(int);
    bign operator-=(const bign&);
    bign operator*(const bign &)const;
    bign operator*(const int num)const;
    bign operator*=(const bign&);
    bign operator/(const bign&)const;
    bign operator/=(const bign&);
//四則運算的衍生運算
    bign operator%(const bign&)const;//取模（餘數）
    bign factorial()const;//階乘
    bign Sqrt()const;//整數開根（向下取整）
    bign pow(const bign&)const;//次方
//輔助的函數
    void clean();
    ~bign();
};
#define max(a,b) a>b ? a : b
#define min(a,b) a<b ? a : b

bign::bign()
{
    memset(s, 0, sizeof(s));
    len = 1;
    sign = 1;
}

bign::bign(const char *num)
{
    *this = num;
}

bign::bign(int num)
{
    *this = num;
}

string bign::toStr() const
{
    string res;
    res = "";
    for (int i = 0; i < len; i++)
        res = (char)(s[i] + '0') + res;
    if (res == "")
        res = "0";
    if (!sign&&res != "0")
        res = "-" + res;
    return res;
}

istream &operator>>(istream &in, bign &num)
{
    string str;
    in>>str;
    num=str;
    return in;
}

ostream &operator<<(ostream &out, bign &num)
{
    out<<num.toStr();
    return out;
}

bign bign::operator=(const char *num)
{
    memset(s, 0, sizeof(s));
    char a[MAX_L] = "";
    if (num[0] != '-')
        strcpy(a, num);
    else
        for (int i = 1; i < strlen(num); i++)
            a[i - 1] = num[i];
    sign = !(num[0] == '-');
    len = strlen(a);
    for (int i = 0; i < strlen(a); i++)
        s[i] = a[len - i - 1] - 48;
    return *this;
}

bign bign::operator=(int num)
{
    if (num < 0)
        sign = 0, num = -num;
    else
        sign = 1;
    char temp[MAX_L];
    sprintf(temp, "%d", num);
    *this = temp;
    return *this;
}

bign bign::operator=(const string num)
{
    const char *tmp;
    tmp = num.c_str();
    *this = tmp;
    return *this;
}

bool bign::operator<(const bign &num) const
{
    if (sign^num.sign)
        return num.sign;
    if (len != num.len)
        return len < num.len;
    for (int i = len - 1; i >= 0; i--)
        if (s[i] != num.s[i])
            return sign ? (s[i] < num.s[i]) : (!(s[i] < num.s[i]));
    return !sign;
}

bool bign::operator>(const bign&num)const
{
    return num < *this;
}

bool bign::operator<=(const bign&num)const
{
    return !(*this>num);
}

bool bign::operator>=(const bign&num)const
{
    return !(*this<num);
}

bool bign::operator!=(const bign&num)const
{
    return *this > num || *this < num;
}

bool bign::operator==(const bign&num)const
{
    return !(num != *this);
}

bign bign::operator+(const bign &num) const
{
    if (sign^num.sign)
    {
        bign tmp = sign ? num : *this;
        tmp.sign = 1;
        return sign ? *this - tmp : num - tmp;
    }
    bign result;
    result.len = 0;
    int temp = 0;
    for (int i = 0; temp || i < (max(len, num.len)); i++)
    {
        int t = s[i] + num.s[i] + temp;
        result.s[result.len++] = t % 10;
        temp = t / 10;
    }
    result.sign = sign;
    return result;
}

bign bign::operator++()
{
    *this = *this + 1;
    return *this;
}

bign bign::operator++(int)
{
    bign old = *this;
    ++(*this);
    return old;
}

bign bign::operator+=(const bign &num)
{
    *this = *this + num;
    return *this;
}

bign bign::operator-(const bign &num) const
{
    bign b=num,a=*this;
    if (!num.sign && !sign)
    {
        b.sign=1;
        a.sign=1;
        return b-a;
    }
    if (!b.sign)
    {
        b.sign=1;
        return a+b;
    }
    if (!a.sign)
    {
        a.sign=1;
        b=bign(0)-(a+b);
        return b;
    }
    if (a<b)
    {
        bign c=(b-a);
        c.sign=false;
        return c;
    }
    bign result;
    result.len = 0;
    for (int i = 0, g = 0; i < a.len; i++)
    {
        int x = a.s[i] - g;
        if (i < b.len) x -= b.s[i];
        if (x >= 0) g = 0;
        else
        {
            g = 1;
            x += 10;
        }
        result.s[result.len++] = x;
    }
    result.clean();
    return result;
}

bign bign::operator * (const bign &num)const
{
    bign result;
    result.len = len + num.len;

    for (int i = 0; i < len; i++)
        for (int j = 0; j < num.len; j++)
            result.s[i + j] += s[i] * num.s[j];

    for (int i = 0; i < result.len; i++)
    {
        result.s[i + 1] += result.s[i] / 10;
        result.s[i] %= 10;
    }
    result.clean();
    result.sign = !(sign^num.sign);
    return result;
}

bign bign::operator*(const int num)const
{
    bign x = num;
    bign z = *this;
    return x*z;
}
bign bign::operator*=(const bign&num)
{
    *this = *this * num;
    return *this;
}

bign bign::operator /(const bign&num)const
{
    bign ans;
    ans.len = len - num.len + 1;
    if (ans.len < 0)
    {
        ans.len = 1;
        return ans;
    }

    bign divisor = *this, divid = num;
    divisor.sign = divid.sign = 1;
    int k = ans.len - 1;
    int j = len - 1;
    while (k >= 0)
    {
        while (divisor.s[j] == 0) j--;
        if (k > j) k = j;
        char z[MAX_L];
        memset(z, 0, sizeof(z));
        for (int i = j; i >= k; i--)
            z[j - i] = divisor.s[i] + '0';
        bign dividend = z;
        if (dividend < divid) { k--; continue; }
        int key = 0;
        while (divid*key <= dividend) key++;
        key--;
        ans.s[k] = key;
        bign temp = divid*key;
        for (int i = 0; i < k; i++)
            temp = temp * 10;
        divisor = divisor - temp;
        k--;
    }
    ans.clean();
    ans.sign = !(sign^num.sign);
    return ans;
}

bign bign::operator/=(const bign&num)
{
    *this = *this / num;
    return *this;
}

bign bign::operator%(const bign& num)const
{
    bign a = *this, b = num;
    a.sign = b.sign = 1;
    bign result, temp = a / b*b;
    result = a - temp;
    result.sign = sign;
    return result;
}

bign bign::pow(const bign& num)const
{
    bign result = 1;
    for (bign i = 0; i < num; i++)
        result = result*(*this);
    return result;
}

bign bign::factorial()const
{
    bign result = 1;
    for (bign i = 1; i <= *this; i++)
        result *= i;
    return result;
}

void bign::clean()
{
    if (len == 0) len++;
    while (len > 1 && s[len - 1] == '\0')
        len--;
}

bign bign::Sqrt()const
{
    if(*this<0)return -1;
    if(*this<=1)return *this;
    bign l=0,r=*this,mid;
    while(r-l>1)
    {
        mid=(l+r)/2;
        if(mid*mid>*this)
            r=mid;
        else 
            l=mid;
    }
    return l;
}

bign::~bign()
{
}</span></span></strong></span>

這是通過數組來儲存大整數，數組的每一個空間儲存一個數字，來達到保存大整數的目的。

大整數加減法的實現跟前面的基本一樣，也是通過模擬進位，借位的方法來實現。

下面來說一下乘法的實現過程，乘法的實現，是通過模擬手算的方法，第一個數的每一位，乘以第二個數的每一位，最後通過彙總，把結果算出來。

而除法則是四則運算中最複雜的一個，基本思路是一個一個減，看最多能減去多少個除數，但顯然這樣做的話效率簡直是極其得低。如何使它減得更快呢？以 7546 除以 23 爲例來看一下：開始商爲 0。先減
去 23 的 100 倍，就是 2300，發現夠減 3 次，餘下 646。於是商的值就增加 300。然後用 646
減去 230，發現夠減 2 次，餘下 186，於是商的值增加 20。最後用 186 減去 23，夠減 8 次，
因此最終商就是 328。

而上面正是用到了這種思路。