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Commit a7d4ca5

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‎3.C++程序设计/week4/编程练习/1.cpp

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‎3.C++程序设计/week4/编程练习/2.cpp

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‎3.C++程序设计/week4/编程练习/3.cpp

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

‎4.算法基础/week2/README.md

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# Week2:枚举
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## 枚举的基本思想
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枚举是基于已有知识进行答案猜测的一种问题求解策略。
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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。
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### 枚举过程
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- 判断猜测的答案是否正确
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- 进行新的猜测: 有两个关键因素要注意
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- 猜测的结果必须是前面的猜测中没有出现过的.
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- 猜测的过程中要及早排除错误的答案.
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### 枚举中的三个关键问题
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问题一:给出解空间
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- 给出解空间,建立简洁的数学模型
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- 模型中变量数尽可能少, 它们之间相互独立
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问题二:减少搜索空间
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- 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
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- 减少代码中循环体执行次数
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问题三:合适的搜索顺序
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- 搜索空间的遍历顺序要与模型中条件表达式一致
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例:求小于N的最大素数
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- 2是素数, 记为PRIM0
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- 根据PRIM0, PRIM1 , ..., PRIMk, 寻找比PRIMk大的最小素数PRIMk+1
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- 如果PRIMk+1大于N, 则PRIMk是我们需要找的素数, 否则继续寻找
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- 解空间:2到N的所有素数
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- 减小搜索空间:排除所有奇数
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