UVa 748 Exponentiation解題報告

高精度小數乘法題，核心是對小數點的處理。

使用string類和bign類可以使代碼比較簡潔。

思路：輸入後進行預處理，把小數點和後面的0去掉。然後進行整數的高精度乘法，最後再按小數的形式進行輸出。

// UVa 748 Exponentiation
#include <iostream>
#include <algorithm>
#include <string>
#include <cstring>
#include <cstdio>
using namespace std;

const int maxn = 10000;
struct bign
{
	int len, s[maxn];

	bign()//構造函數
	{
		memset(s, 0, sizeof(s));
		len = 1;
	}

	bign operator = (const char * num)//重載運算符=
	{
		len = strlen(num);
		for (int i = 0; i < len; i++)
			s[i] = num[len - i - 1] - '0';
		return *this;
	}

	bign(const char * num){ *this = num; }//支援初始化操作

	bign operator = (int num)
	{
		char s[maxn];
		sprintf(s, "%d", num);//把num輸出到字元串s中
		*this = s;
		return *this;
	}

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

	string str() const     //利用string類把字元串數組轉換為字元串，友善用<<,>>輸出
	{
		string res = "";
		for (int i = 0; i < len; i++)
			res = (char)(s[i] + '0') + res;
		if (res == "")	res = "0";
		return res;
	}

	bign operator + (const bign& b) const//定義加法
	{
		bign c;
		c.len = 0;
		for (int i = 0, g = 0; g || i < max(len, b.len); i++)//要兩個數的位數都計算一便
		{
			int x = g;//g為其餘數
			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;
	}

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

	void clean()//把0排除，得到真實的len
	{
		while (len > 1 && !s[len - 1])
			len--;
	}

	bign operator - (const bign& b) const
	{
		bign c;
		c.len = 0;
		for (int i = 0, g = 0; i < len; i++)
		{
			int x = s[i] - g;//減去借1
			if (i < b.len)	x -= b.s[i];//上減下
			if (x >= 0)		g = 0;
			else
			{
				g = 1;//x<0說明需要向前借1
				x += 10;//将x變為正
			}
			c.s[c.len++] = x;
		}
		c.clean();//x可能為0
		return c;
	}

	bign operator -= (const bign& b)
	{
		*this = *this - b;
		return *this;
	}

	bign operator * (const bign& b) const
	{
		bign c;
		c.len = len + b.len;//結果的位數最大為兩個因子位數之和
		for (int i = 0; i < len; i++)
		for (int j = 0; j < b.len; j++)
			c.s[i + j] += s[i] * b.s[j];//對應位置的積的和，累加起來就是結果
		for (int i = 0; i < c.len - 1; i++)
		{
			c.s[i + 1] += c.s[i] / 10;//進位的值
			c.s[i] %= 10;//餘數位
		}
		c.clean();
		return c;
	}

	bign operator *= (const bign& b)
	{
		*this = *this * b;
		return *this;
	}
	//重載比較運算符
	bool operator < (const bign& b) const
	{
		if (len != b.len)
			return len < b.len;
		for (int i = len - 1; i >= 0; i--)
		if (s[i] != b.s[i])
			return len < b.len;
		return false;
	}

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

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

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

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

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

//重載<<,>>，支援直接輸出bign對象
istream& operator >> (istream &in, bign& x)//
{
	string s;
	in >> s;
	x = s.c_str();
	return in;
}

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

int main()
{
	//freopen("data.txt", "r", stdin);
	string a, b;
	int n;
	while (cin >> a >> n)
	{
		int i = a.find('.');//找出小數點的位置

		for(int i = a.length() - 1; i >= 0; i--)//除去後面的0
			if(a[i] == '0')
				a.erase(i);
			else
				break;
		b = a.substr(0, i) + a.substr(i + 1);//把小數點去掉組成新的數

		const char * str = b.c_str();//轉化為c字元串
		bign num = str;//轉化為bign對象
		bign product = str;
		for(int j = 1; j < n; j++)//累乘
			product *= num;
		int dos = (b.length() - i) * n;//得到小數點到最後非零位的長度
		string c = product.str();//把bign對象轉化為string對象
		dos = c.length() - dos;//得到小數點的新位置
		if(dos < 0)//說明沒有整數部分
		{
			printf(".");
			for(int k = dos; k < 0; k++)//把缺少的前導0補充輸出
				printf("0");
		}

		for(int j = 0; j < c.length(); j++)//輸出c
			if(j == dos)
			{
				printf(".");
				j--;//找到小數點的位置後要j--，使c繼續不間斷輸出
				dos = -1;//防止無限循環，缺少這步會使條件一直成立。
			}
			else
				cout << c[j];
		cout << endl;
	}
}