大数相加 大数相乘 C++

最简单的思路是直接用int型数组存储大数的每一位，程序比较容易实现，但是效率稍低，直接上代码。

#include
#include
using namespace std;
class bign
{
	int num[1000];
	int pos_high;//最高位所在位置
public:
	bign() { pos_high = -1; }
	bign(const bign&t)
	{
		memcpy(num, t.num, sizeof(num));
		pos_high = t.pos_high;
	}
	bign operator=(const bign&t)
	{
		memcpy(num, t.num, sizeof(num));
		pos_high = t.pos_high;
		return t;
	}
	bign operator=(const char*t)
	{
		int len = strlen(t);
		for (int i = 0;i < len;i++)
			num[i] = t[len - i - 1]-'0';
		pos_high = len - 1;
		return *this;
	}
	bign operator+(const bign&t)const
	{
		bign ans;
		int i,r = 0,s;
		for ( i = 0;i <= pos_high || i <= t.pos_high;i++)
		{
			if (i > t.pos_high)
			{
				s = num[i] + r;
				ans.num[i] = s%10;
				r = s / 10;
			}
			else if (i > pos_high)
			{
				s = t.num[i] + r;
				ans.num[i] = s % 10;
				r = s / 10;
			}
			else
			{
				s = num[i] + t.num[i]+r;
				ans.num[i]=s % 10;
				r = s / 10;
			}
		}
		if (r > 0)
			ans.num[i] = 1, ans.pos_high = i;
		else
			ans.pos_high = i - 1;
		return ans;
	}
	bign operator*(const bign&t)const
	{
		bign te,ans;
		int i, j,r,s;
		char str[1000];
		for (i = 0;i <= t.pos_high;i++)
		{
			r = 0;
			te.pos_high = 0;
			for (int k = 0;k < i;k++)
				str[k] = '0';
			str[i] = '\0';
			te = str;
			for (j = 0;j <= pos_high;j++)
			{
				s = t.num[i]*num[j] + r;
				te.num[j+i] =s% 10;
				r = s / 10;
			}
			if (r > 0)
				te.num[j+i] = r,te.pos_high=j+i;
			else
				 te.pos_high = j-1+i;
			ans = ans + te;
		}
		return ans;
	}
	bign operator++()
	{
		int i,r=1,s;
		for (i = 0;i <= pos_high&&r>0;i++)
		{
			s = num[i] + r;
			num[i] = s % 10;
			r = s / 10;
		}
		if (r > 0)
			num[i] = 1,pos_high++;
		return *this;
	}
	bool operator<(const bign&t)
	{
		int i;
		if (t.pos_high != pos_high)
			return pos_high < t.pos_high;
		else
		{
			for ( i = pos_high;i >= 0 && num[i] == t.num[i];i--);
			return num[i] < t.num[i];
		}
	}
	friend ostream& operator<<(ostream&out,const bign&x)
	{
		for (int i = x.pos_high;i >= 0;i--)
			out << x.num[i];
		return out;
	}
	int mod(int MOD)
	{
		int i, res = 0;
		for (i = pos_high;i>=0;i--)
			res = (num[i] + res * 10)%MOD;
		return res;
	}
};
int main()
{
	bign a,b,c,d;
         a = "9999999999999999999999999999999999999999999999999999999999999";
	 b = "9999999999999999999999999999999999999999999999999999999999999999999";
	c = a + b;
	d = a*b;
	cout << c << endl;
	cout << d << endl;
	
	return 0;
	cout << n;

}

int型可以表示所有9位数，仅表示一位不仅浪费空间，而且增加了9倍的运算次数，故可以用int表示9位数进行优化，如123456789这个数占1个，1234567890占两个。乘法采用模拟手动算法的方法，各位依次相乘，最后相加。直接看代码吧。

#include
#include
using namespace std;
const int MOD = 1000000000;
class bign
{
	int num[10000];//10^9进制
	int len;      //大数长度
public:
	bign() { len = 0; }
	bign(char*t)
	{
		int length = strlen(t);
		int a, b, sum,c=0;
		a = length / 9;
		b = length % 9;
		for (int i = 0;i < a;i++)
		{
			sum = 0;
			for (int j = 0;j<9 ;j++)
				sum = sum * 10 +( t[b+j + (a-i-1) * 9]-'0');
			num[i] = sum;
		}
		if (b > 0){
			sum = 0;
			for (int i = 0;i  0)
			num[len] = 1, len = len + 1;
		return*this;
	}
	bign operator+(const bign&t)const
	{
		bign ans;
		int  r = 0, s, i;
		for (i = 0;i < len || i < t.len;i++)
		{
			if (i >= len)
			{
				s = t.num[i] + r;
				ans.num[i] = s % MOD;
				r = s / MOD;
			}
			else if (i >= t.len)
			{
				s = num[i] + r;
				ans.num[i] = s % MOD;
				r = s / MOD;
			}
			else
			{
				s = num[i] + t.num[i] + r;
				ans.num[i] = s % MOD;
				r = s / MOD;
			}
		}
		if (r > 0)
			ans.num[i] = r, ans.len = i + 1;
		else
			ans.len = i;
		return ans;
	}
	bign operator+(const int&t)
	{
		bign ans;
		int r =t, s,i;
		for (i = 0;i < len&&r>0;i++)
		{
			s = num[i] + r;
			ans.num[i] = s % MOD;
			r = s / MOD;
		}
		if (r > 0)
			ans.num[len] = 1, ans.len = len + 1;
		else
		{
			for (;i < len;i++)
				ans.num[i] = num[i];
			ans.len = len;
		}
		return ans;
	}
	bign operator*(const bign&t)const
	{
		bign ans, te;
		int   i,j;
		long long r = 0, s;
		for (i = 0;i < len;i++ )
		{
			if (num[i] == 0) continue;
			for (int k = 0;k < i;k++)
				te.num[k] = 0;
			for (j = 0;j < t.len;j++)
			{
				s =(long long)num[i] * (long long)t.num[j]+r;
				r = s /MOD;
				te.num[i + j] =(int)( s % MOD);
			}
			if (r > 0)
				te.num[i + j] = (int)r, te.len = i + j + 1,r=0;
			else
				te.len=i + j;
			ans = ans + te;
		}
		return ans;
	}
	bign operator*(const int&t)
	{
		bign ans;
		long long r = 0, s;
		int i;
		for (i = 0;i < len;i++)
		{
			s = (long long)num[i] * (long long)t + r;
			r = s / MOD;
			ans.num[i] = (int)(s % MOD);
		}
		if (r > 0)
			ans.num[i] = (int)r, ans.len = i + 1, r = 0;
		else
			ans.len = i ;
		return ans;
	}
	friend ostream& operator<<(ostream&out, const bign&t)
	{
		int i = t.len - 1;
		if (i >= 0) {
			out << t.num[i];//最高位直接输出
			for (i--;i >= 0;i--)
			{
				out.width(9);//设置位宽
				out.fill('0');//补充前置0，如123输出000000123
				out << t.num[i];
			}
		}
		else out << 0;//大数占0位直接输出0
		return out;
	}
	bool operator<(const bign&t)
	{
		if (t.len != len)
			return len < t.len;
		else
		{
			int i;
			for (i = len - 1;i >= 0 && t.num[i] == num[i];i--);
			return num[i] < t.num[i];
		}
	}
};
int main()
{
	bign a = "9999999999999999999999999999999999999999999999999999999999999",
		b = "9999999999999999999999999999999999999999999999999999999999999999999", c,d;
	c = a + b;
	d = a*b;
	cout << c << endl;
	cout << d << endl;
	return 0;

}