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Author: chiuchiuuu

3.C++程序设计/week4/编程练习/1.cpp

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+#include <iostream>
+#include <string>
+#include <cstdlib>
+using namespace std;
+
+class BigInt
+{
+private:
+	string num;
+	char sign;
+public:
+	BigInt(string n = "0");
+	BigInt operator+(const BigInt &bi);
+	BigInt operator-(const BigInt &bi);
+	BigInt operator*(const BigInt &bi);
+	BigInt operator/(const BigInt &bi);
+	friend ostream& operator<<(ostream &o, const BigInt &b);
+};
+
+BigInt::BigInt(string n)
+{
+	if (n[0] == '-')
+	{
+		sign = '-';
+		num = n.substr(1, n.length() - 1);
+	}
+	else
+	{
+		sign = '+';
+		num = n;
+	}
+}
+
+BigInt BigInt::operator+(const BigInt &bi)
+{
+	string operand1 = num;
+	string operand2 = bi.num;
+
+	int maxLength = operand1.length() > operand2.length() ? operand1.length() : operand2.length();
+	while (operand2.length() < operand1.length())
+	{
+		operand2.insert(0, "0");
+	}
+	while (operand1.length() < operand2.length())
+	{
+		operand1.insert(0, "0");
+	}
+
+	char *result = new char[maxLength + 1];
+	result[maxLength] = '\0';
+	int carry = 0;
+	for (int i = maxLength - 1; i >= 0; i--)
+	{
+		int add1 = operand1[i] - '0';
+		int add2 = operand2[i] - '0';
+		int add = (add1 + add2 + carry) % 10;
+		result[i] = '0' + add;
+		carry = (add1 + add2 + carry) / 10;
+	}
+
+	BigInt tmp(result);
+	if (carry)
+	{
+		tmp.num = "1" + tmp.num;
+	}
+	tmp.sign = sign;
+	return tmp;;
+}
+
+BigInt BigInt::operator-(const BigInt &bi)
+{
+	string operand1 = num;
+	string operand2 = bi.num;
+	BigInt tmp;
+	if (operand1 == operand2)
+	{
+		tmp.num = "0";
+		tmp.sign = '+';
+		return tmp;
+	}
+
+	int maxLength = 0;
+	char tmpSign = '+';
+	if (operand1.length() < operand2.length() || (operand1.length() == operand2.length() && operand1 < operand2))
+	{
+		tmpSign = '-';
+		swap(operand1, operand2);
+	}
+	maxLength = operand1.length();
+	while (operand2.length() < operand1.length())
+	{
+		operand2.insert(0, "0");
+	}
+
+	char *result = new char[maxLength + 1];
+	result[maxLength] = '\0';
+	int borrow = 0;
+	for (int i = maxLength - 1; i >= 0; i--)
+	{
+		int minus1 = operand1[i] - '0';
+		int minus2 = operand2[i] - '0';
+		int minus = minus1 - minus2 - borrow;
+		if (minus < 0)
+		{
+			minus += 10;
+			borrow = 1;
+		}
+		else
+		{
+			borrow = 0;
+		}
+		result[i] = '0' + minus;
+	}
+
+	tmp.num = result;
+	while (tmp.num[0] == '0')
+	{
+		tmp.num.erase(tmp.num.begin());
+	}
+	tmp.sign = tmpSign;
+	return tmp;
+}
+
+BigInt BigInt::operator*(const BigInt &bi)
+{
+	string operand1 = num;
+	string operand2 = bi.num;
+	BigInt tmp;
+	BigInt result;
+	result.sign = '+';
+
+	for (int i = 0; i < operand2.length(); i++)
+	{
+		tmp.num = "0";
+		for (int j = 0; j < operand2[i] - '0'; j++)
+		{
+			tmp = tmp + *this;
+		}
+		tmp.num.resize(tmp.num.length() + i, '0');
+		result = result + tmp;
+	}
+	return result;
+}
+
+BigInt BigInt::operator/(const BigInt &bi)
+{
+	string operand1 = num;
+	string operand2 = bi.num;
+	BigInt quotient;
+	quotient.sign = '+';
+	if (operand1.length() < operand2.length())
+	{
+		quotient.num = "0";
+		return quotient;
+	}
+
+	int m = operand1.length() - operand2.length();
+	operand2.resize(operand1.length(), '0');
+	BigInt dividend(operand1);
+	BigInt divisor(operand2);
+	for (int i = 0; i <= m; i++)
+	{
+		int tmp = 0;
+		while ((dividend - divisor).sign != '-')
+		{
+			tmp++;
+			dividend = dividend - divisor;
+		}
+		quotient.num += ('0' + tmp);
+		divisor.num.resize(divisor.num.length() - 1);
+	}
+	while (quotient.num[0] == '0')
+	{
+		quotient.num.erase(quotient.num.begin());
+	}
+	return quotient;
+}
+
+
+
+ostream& operator<<(ostream &o, const BigInt &b)
+{
+	if (b.sign == '+')
+	{
+		o << b.num;
+	}
+	else if (b.sign == '-')
+	{
+		o << "-" << b.num;
+	}
+	return o;
+}
+
+int main()
+{
+	string s1, s2;
+	char Operator;
+	cin >> s1 >> Operator >> s2;
+	BigInt a(s1);
+	BigInt b(s2);
+	BigInt result;
+
+	switch (Operator)
+	{
+	case '+':
+		result = a + b;
+		break;
+	case '-':
+		result = a - b;
+		break;
+	case '*':
+		result = a * b;
+		break;
+	case '/':
+		result = a / b;
+		break;
+	}
+	cout << result << endl;
+	return 0;
+}

3.C++程序设计/week4/编程练习/2.cpp

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+# Week2：枚举
+
+## 枚举的基本思想
+
+枚举是基于已有知识进行答案猜测的一种问题求解策略。
+
+枚举从可能的集合中一一列举各元素， 对问题可能解集合的每一项，根据问题给定的检验条件判定哪些是成立的，使条件成立的即是问题的解。
+
+### 枚举过程
+
+- 判断猜测的答案是否正确
+- 进行新的猜测: 有两个关键因素要注意
+  - 猜测的结果必须是前面的猜测中没有出现过的. 
+  - 猜测的过程中要及早排除错误的答案. 
+
+### 枚举中的三个关键问题
+
+问题一：给出解空间
+
+- 给出解空间，建立简洁的数学模型 
+- 模型中变量数尽可能少, 它们之间相互独立 
+
+问题二：减少搜索空间
+
+- 利用知识缩小模型中各变量的取值范围, 避免不必要的计算 
+- 减少代码中循环体执行次数
+
+问题三：合适的搜索顺序
+
+- 搜索空间的遍历顺序要与模型中条件表达式一致 
+
+例：求小于N的最大素数
+
+- 2是素数, 记为PRIM0
+- 根据PRIM0, PRIM1 , …, PRIMk, 寻找比PRIMk大的最小素数PRIMk+1
+- 如果PRIMk+1大于N, 则PRIMk是我们需要找的素数, 否则继续寻找 
+- 解空间：2到N的所有素数
+- 减小搜索空间：排除所有奇数