大数模版

#include <stdio.h>
#include <string.h> 
#include <stdlib.h> 
#include <math.h>
#include <assert.h>  
#include <ctype.h> 
#include <map>
#include <string>
#include <set>
#include <bitset>
#include <utility>
#include <algorithm>
#include <vector>
#include <stack>
#include <queue>
#include <iostream>
#include <fstream>
#include <list>
using  namespace  std;      
     
const  int MAXL = 500;      
struct  BigNum      
{      
    int  num[MAXL];      
    int  len;      
};      
     
//高精度比较 a > b return 1, a == b return 0; a < b return -1;      
int  Comp(BigNum &a, BigNum &b)      
{      
    int  i;      
    if(a.len != b.len) return (a.len > b.len) ? 1 : -1;      
    for(i = a.len-1; i >= 0; i--)      
        if(a.num[i] != b.num[i]) return  (a.num[i] > b.num[i]) ? 1 : -1;      
    return  0;      
}      
     
//高精度加法      
BigNum  Add(BigNum &a, BigNum &b)      
{      
    BigNum c;      
    int  i, len;      
    len = (a.len > b.len) ? a.len : b.len;      
    memset(c.num, 0, sizeof(c.num));      
    for(i = 0; i < len; i++)      
    {      
        c.num[i] += (a.num[i]+b.num[i]);      
        if(c.num[i] >= 10)      
        {      
            c.num[i+1]++;      
            c.num[i] -= 10;      
        }      
    }      
    if(c.num[len])
		len++;      
    c.len = len;      
    return  c;      
}      
//高精度减法,保证a >= b      
BigNum Sub(BigNum &a, BigNum &b)      
{      
    BigNum  c;      
    int  i, len;      
    len = (a.len > b.len) ? a.len : b.len;      
    memset(c.num, 0, sizeof(c.num));      
    for(i = 0; i < len; i++)      
    {      
        c.num[i] += (a.num[i]-b.num[i]);      
        if(c.num[i] < 0)      
        {      
            c.num[i] += 10;      
            c.num[i+1]--;      
        }      
    }      
    while(c.num[len] == 0 && len > 1)
		len--;      
    c.len = len;      
    return  c;      
}      
//高精度乘以低精度，当b很大时可能会发生溢出int范围,具体情况具体分析      
//如果b很大可以考虑把b看成高精度      
BigNum Mul1(BigNum &a, int  &b)      
{      
    BigNum c;      
    int  i, len;      
    len = a.len;      
    memset(c.num, 0, sizeof(c.num));      
    //乘以0，直接返回0      
    if(b == 0)       
    {      
        c.len = 1;      
        return  c;      
    }      
    for(i = 0; i < len; i++)      
    {      
        c.num[i] += (a.num[i]*b);      
        if(c.num[i] >= 10)      
        {      
            c.num[i+1] = c.num[i]/10;      
            c.num[i] %= 10;      
        }      
    }      
    while(c.num[len] > 0)      
    {      
        c.num[len+1] = c.num[len]/10;      
        c.num[len++] %= 10;      
    }      
    c.len = len;       
    return  c;      
}      
     
//高精度乘以高精度,注意要及时进位，否则肯能会引起溢出,但这样会增加算法的复杂度，      
//如果确定不会发生溢出, 可以将里面的while改成if      
BigNum  Mul2(BigNum &a, BigNum &b)      
{      
    int i, j, len = 0;      
    BigNum  c;      
    memset(c.num, 0, sizeof(c.num));      
    for(i = 0; i < a.len; i++)
	{
        for(j = 0; j < b.len; j++)      
        {      
            c.num[i+j] += (a.num[i]*b.num[j]);      
            if(c.num[i+j] >= 10)      
            {      
                c.num[i+j+1] += c.num[i+j]/10;      
                c.num[i+j] %= 10;      
            }      
        }
	}
    len = a.len+b.len-1;      
    while(c.num[len-1] == 0 && len > 1)
		len--;      
    if(c.num[len])
		len++;      
    c.len = len;      
    return  c;      
}      
     
//高精度除以低精度,除的结果为c, 余数为f      
void Div1(BigNum &a, int &b, BigNum &c, int &f)      
{      
    int  i, len = a.len;      
    memset(c.num, 0, sizeof(c.num));      
    f = 0;      
    for(i = a.len-1; i >= 0; i--)      
    {      
        f = f*10+a.num[i];      
        c.num[i] = f/b;      
        f %= b;      
    }      
    while(len > 1 && c.num[len-1] == 0)
		len--;      
    c.len = len;      
}      
//高精度*10      
void  Mul10(BigNum &a)      
{      
    int  i, len = a.len;      
    for(i = len; i >= 1; i--)      
        a.num[i] = a.num[i-1];      
    a.num[i] = 0;      
    len++;      
    //if a == 0      
    while(len > 1 && a.num[len-1] == 0)
		len--;      
}      
     
//高精度除以高精度，除的结果为c，余数为f      
void Div2(BigNum &a, BigNum &b, BigNum &c, BigNum &f)      
{      
    int  i, len = a.len;      
    memset(c.num, 0, sizeof(c.num));      
    memset(f.num, 0, sizeof(f.num));      
    f.len = 1;      
    for(i = len-1;i >= 0;i--)      
    {      
        Mul10(f);      
        //余数每次乘10      
        f.num[0] = a.num[i];      
        //然后余数加上下一位      
        ///利用减法替换除法      
        while(Comp(f, b) >= 0)      
        {
            f = Sub(f, b);      
            c.num[i]++;      
        }      
    }      
    while(len > 1 && c.num[len-1] == 0)
		len--;      
    c.len = len;      
}   
void  print(BigNum &a)   //输出大数   
{      
    int  i;      
    for(i = a.len-1; i >= 0; i--)      
        printf("%d", a.num[i]);      
    puts("");      
}      
//将字符串转为大数存在BigNum结构体里面      
BigNum ToNum(char *s)      
{      
    int i, j;      
    BigNum  a;      
    a.len = strlen(s);      
    for(i = 0, j = a.len-1; s[i] != '\0'; i++, j--)      
        a.num[i] = s[j]-'0';      
    return  a;      
}      
     
void Init(BigNum &a, char *s, int &tag)   //将字符串转化为大数
{   
    int  i = 0, j = strlen(s); 
    if(s[0] == '-')
	{
		j--;
		i++;
		tag *= -1;
	}
    a.len = j;
    for(; s[i] != '\0'; i++, j--)
        a.num[j-1] = s[i]-'0';
}   
  
int main(void)      
{      
    BigNum a, b;   
    char  s1[100], s2[100];   
    while(scanf("%s %s", s1, s2) != EOF)   
    {   
        int tag = 1;   
        Init(a, s1, tag);    //将字符串转化为大数
        Init(b, s2, tag);   
        a = Mul2(a, b);   
        if(a.len == 1 && a.num[0] == 0)   
        {   
            puts("0");   
        }   
        else    
        {   
            if(tag < 0) putchar('-');   
            print(a);   
        }   
    }   
    return 0;   
}

#include<iostream> 
#include<string> 
#include<iomanip> 
#include<algorithm> 
using namespace std; 
#pragma warning(disable : 4996)
#define MAXN 9999
#define MAXSIZE 10
#define DLEN 4

class BigNum
{ 
private: 
	int a[500];    //可以控制大数的位数 
	int len;       //大数长度
public: 
	BigNum(){ len = 1;memset(a,0,sizeof(a)); }   //构造函数
	BigNum(const int);       //将一个int类型的变量转化为大数
	BigNum(const char*);     //将一个字符串类型的变量转化为大数
	BigNum(const BigNum &);  //拷贝构造函数
	BigNum &operator=(const BigNum &);   //重载赋值运算符，大数之间进行赋值运算

	friend istream& operator>>(istream&,  BigNum&);   //重载输入运算符
	friend ostream& operator<<(ostream&,  BigNum&);   //重载输出运算符

	BigNum operator+(const BigNum &) const;   //重载加法运算符，两个大数之间的相加运算 
	BigNum operator-(const BigNum &) const;   //重载减法运算符，两个大数之间的相减运算 
	BigNum operator*(const BigNum &) const;   //重载乘法运算符，两个大数之间的相乘运算 
	BigNum operator/(const int   &) const;    //重载除法运算符，大数对一个整数进行相除运算

	BigNum operator^(const int  &) const;    //大数的n次方运算
	int    operator%(const int  &) const;    //大数对一个int类型的变量进行取模运算    
	bool   operator>(const BigNum & T)const;   //大数和另一个大数的大小比较
	bool   operator>(const int & t)const;      //大数和一个int类型的变量的大小比较

	void print();       //输出大数
}; 
BigNum::BigNum(const int b)     //将一个int类型的变量转化为大数
{ 
	int c,d = b;
	len = 0;
	memset(a,0,sizeof(a));
	while(d > MAXN)
	{
		c = d - (d / (MAXN + 1)) * (MAXN + 1); 
		d = d / (MAXN + 1);
		a[len++] = c;
	}
	a[len++] = d;
}
BigNum::BigNum(const char*s)     //将一个字符串类型的变量转化为大数
{
	int t,k,index,l,i;
	memset(a,0,sizeof(a));
	l=strlen(s);   
	len=l/DLEN;
	if(l%DLEN)
		len++;
	index=0;
	for(i=l-1;i>=0;i-=DLEN)
	{
		t=0;
		k=i-DLEN+1;
		if(k<0)
			k=0;
		for(int j=k;j<=i;j++)
			t=t*10+s[j]-'0';
		a[index++]=t;
	}
}
BigNum::BigNum(const BigNum & T) : len(T.len)  //拷贝构造函数
{ 
	int i; 
	memset(a,0,sizeof(a)); 
	for(i = 0 ; i < len ; i++)
		a[i] = T.a[i]; 
} 
BigNum & BigNum::operator=(const BigNum & n)   //重载赋值运算符，大数之间进行赋值运算
{
	int i;
	len = n.len;
	memset(a,0,sizeof(a)); 
	for(i = 0 ; i < len ; i++) 
		a[i] = n.a[i]; 
	return *this; 
}
istream& operator>>(istream & in,  BigNum & b)   //重载输入运算符
{
	char ch[MAXSIZE*4];
	int i = -1;
	in>>ch;
	int l=strlen(ch);
	int count=0,sum=0;
	for(i=l-1;i>=0;)
	{
		sum = 0;
		int t=1;
		for(int j=0;j<4&&i>=0;j++,i--,t*=10)
		{
			sum+=(ch[i]-'0')*t;
		}
		b.a[count]=sum;
		count++;
	}
	b.len =count++;
	return in;

}
ostream& operator<<(ostream& out,  BigNum& b)   //重载输出运算符
{
	int i;  
	cout << b.a[b.len - 1]; 
	for(i = b.len - 2 ; i >= 0 ; i--)
	{ 
		cout.width(DLEN); 
		cout.fill('0'); 
		cout << b.a[i]; 
	} 
	return out;
}

BigNum BigNum::operator+(const BigNum & T) const   //两个大数之间的相加运算
{
	BigNum t(*this);
	int i,big;      //位数   
	big = T.len > len ? T.len : len; 
	for(i = 0 ; i < big ; i++) 
	{ 
		t.a[i] +=T.a[i]; 
		if(t.a[i] > MAXN) 
		{ 
			t.a[i + 1]++; 
			t.a[i] -=MAXN+1; 
		} 
	} 
	if(t.a[big] != 0)
		t.len = big + 1; 
	else
		t.len = big;   
	return t;
}
BigNum BigNum::operator-(const BigNum & T) const   //两个大数之间的相减运算 
{  
	int i,j,big;
	bool flag;
	BigNum t1,t2;
	if(*this>T)
	{
		t1=*this;
		t2=T;
		flag=0;
	}
	else
	{
		t1=T;
		t2=*this;
		flag=1;
	}
	big=t1.len;
	for(i = 0 ; i < big ; i++)
	{
		if(t1.a[i] < t2.a[i])
		{ 
			j = i + 1; 
			while(t1.a[j] == 0)
				j++; 
			t1.a[j--]--; 
			while(j > i)
				t1.a[j--] += MAXN;
			t1.a[i] += MAXN + 1 - t2.a[i]; 
		} 
		else
			t1.a[i] -= t2.a[i];
	}
	t1.len = big;
	while(t1.a[len - 1] == 0 && t1.len > 1)
	{
		t1.len--; 
		big--;
	}
	if(flag)
		t1.a[big-1]=0-t1.a[big-1];
	return t1; 
} 

BigNum BigNum::operator*(const BigNum & T) const   //两个大数之间的相乘运算 
{ 
	BigNum ret; 
	int i,j,up; 
	int temp,temp1;   
	for(i = 0 ; i < len ; i++)
	{ 
		up = 0; 
		for(j = 0 ; j < T.len ; j++)
		{ 
			temp = a[i] * T.a[j] + ret.a[i + j] + up; 
			if(temp > MAXN)
			{ 
				temp1 = temp - temp / (MAXN + 1) * (MAXN + 1); 
				up = temp / (MAXN + 1); 
				ret.a[i + j] = temp1; 
			} 
			else
			{ 
				up = 0; 
				ret.a[i + j] = temp; 
			} 
		} 
		if(up != 0) 
			ret.a[i + j] = up; 
	} 
	ret.len = i + j; 
	while(ret.a[ret.len - 1] == 0 && ret.len > 1)
		ret.len--; 
	return ret; 
} 
BigNum BigNum::operator/(const int & b) const   //大数对一个整数进行相除运算
{ 
	BigNum ret; 
	int i,down = 0;   
	for(i = len - 1 ; i >= 0 ; i--)
	{ 
		ret.a[i] = (a[i] + down * (MAXN + 1)) / b; 
		down = a[i] + down * (MAXN + 1) - ret.a[i] * b; 
	} 
	ret.len = len; 
	while(ret.a[ret.len - 1] == 0 && ret.len > 1)
		ret.len--; 
	return ret; 
}
int BigNum::operator %(const int & b) const    //大数对一个int类型的变量进行取模运算    
{
	int i,d=0;
	for (i = len-1; i>=0; i--)
	{
		d = ((d * (MAXN+1))% b + a[i])% b;  
	}
	return d;
}
BigNum BigNum::operator^(const int & n) const    //大数的n次方运算
{
	BigNum t,ret(1);
	int i;
	if(n<0)
		exit(-1);
	if(n==0)
		return 1;
	if(n==1)
		return *this;
	int m=n;
	while(m>1)
	{
		t=*this;
		for( i=1;i<<1<=m;i<<=1)
		{
			t=t*t;
		}
		m-=i;
		ret=ret*t;
		if(m==1)
			ret=ret*(*this);
	}
	return ret;
}
bool BigNum::operator>(const BigNum & T) const   //大数和另一个大数的大小比较
{ 
	int ln; 
	if(len > T.len)
		return true; 
	else if(len == T.len)
	{ 
		ln = len - 1; 
		while(a[ln] == T.a[ln] && ln >= 0)
			ln--; 
		if(ln >= 0 && a[ln] > T.a[ln])
			return true; 
		else
			return false; 
	} 
	else
		return false; 
}
bool BigNum::operator >(const int & t) const    //大数和一个int类型的变量的大小比较
{
	BigNum b(t);
	return *this>b;
}

void BigNum::print()    //输出大数
{ 
	int i;   
	cout << a[len - 1]; 
	for(i = len - 2 ; i >= 0 ; i--)
	{ 
		cout.width(DLEN); 
		cout.fill('0'); 
		cout << a[i]; 
	} 
	cout << endl;
}
int main(void)
{
	char str1[MAXN], str2[MAXN];
	scanf("%s %s", str1, str2);
	BigNum text1(str1);
	BigNum text2(str2);
	BigNum text3 = text1 + text2;
	text3.print();
}