现在的位置: 首页 > 综合 > 正文

UVA 10106 Product

2019年04月08日 ⁄ 综合 ⁄ 共 1572字 ⁄ 字号 评论关闭

练习bign类高精度。

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <string>
#include <algorithm>
#include <climits>
using namespace std;

const int MAXN = 11000;
const int INF = INT_MAX;

struct bign
{
	int len, s[MAXN];
	bign ()
	{
		memset(s, 0, sizeof(s));
		len = 1;
	}
	bign (int num) {*this = num;}
	bign (const char *num) { *this = num ;}
	bign operator = (const char *num)
	{
		len = strlen(num);
		for(int i = 0; i < len; i++) s[i] = num[len-i-1] - '0';
		return *this;
	}
	bign operator = (int num)
	{
		char s[MAXN];
		sprintf(s, "%d", num);
		*this = s;
		return *this;
	}
	bign operator + (const bign &b) const
	{
		bign c;
		c.len = 0;
		for(int i = 0, g = 0; g || i < max(len, b.len); i++)
		{
			int x = g;
			if(i < len) x += s[i];
			if(i < b.len) x += b.s[i];
			c.s[c.len++] = x % 10;
			g = x / 10;
		}
		return c;
	}
	void clean()
	{
		while(len > 1 && !s[len-1]) len--;
	}
	bign operator * (const bign &b)
	{
		bign c;
		c.len = len + b.len;
		for(int i = 0; i < len; i++)
		{
			for(int j = 0; j < b.len; j++)
			{
				c.s[i+j] += s[i] * b.s[j];
			}
		}
		for(int i = 0; i < c.len; i++)
		{
			c.s[i+1] += c.s[i] / 10;
			c.s[i] %= 10;
		}
		c.clean();
		return c;
	}
	bign operator += (const bign &b)
	{
		*this = *this + b;
		return *this;
	}
	bool operator < (const bign &b)
	{
		if(len != b.len) return len < b.len;
		for(int i = len-1; i >= 0; i--)
		{
			if(s[i] != b.s[i]) return s[i] < b.s[i];
		}
		return false;
	}
	bool operator > (const bign &b)
	{
		if(len != b.len) return len > b.len;
		for(int i = len-1; i >= 0; i--)
		{
			if(s[i] != b.s[i]) return s[i] > b.s[i];
		}
		return false;
	}
	bool operator == (const bign &b)
	{
		return !(*this > b) && !(*this < b);
	}
	string str() const
	{
		string res = "";
		for(int i = 0; i < len; i++) res = char(s[i]+'0') + res;
		if(res == "") res = "0";
		return res;
	}
};

istream& operator >> (istream &in, bign &x)
{
	string s;
	in >> s;
	x = s.c_str();
	return in;
}

ostream& operator << (ostream &out, bign &x)
{
	out << x.str();
	return out;
}

int main()
{
	bign a, b, ans;
	while(cin>>a>>b)
	{
		ans = a*b;
		cout<<ans<<endl;
	}
	return 0;
}

抱歉!评论已关闭.