💄 Optimize styling

2025-09-13 01:20:09 +08:00
parent 4cf035f2f1
commit a9a1680f5d
4 changed files with 163 additions and 71 deletions
--- a/lib/solver.dart
+++ b/lib/solver.dart
@@ -0,0 +1,576 @@
+import 'dart:math';
+import 'package:flutter/foundation.dart'; // For kDebugMode
+import 'package:math_expressions/math_expressions.dart';
+import 'models/calculation_step.dart';
+
+/// 帮助解析一元一次方程 ax+b=cx+d 的辅助类
+class LinearEquationParts {
+  final double a, b, c, d;
+  LinearEquationParts(this.a, this.b, this.c, this.d);
+}
+
+class SolverService {
+  /// 主入口方法，识别并分发任务
+  CalculationResult solve(String input) {
+    // 预处理输入字符串
+    final cleanInput = input.replaceAll(' ', '').toLowerCase();
+
+    // 对包含x的方程进行预处理，展开表达式
+    String processedInput = cleanInput;
+    if (processedInput.contains('x') && processedInput.contains('(')) {
+      processedInput = _expandExpressions(processedInput);
+    }
+
+    // 1. 检查是否为二元一次方程组 (格式: ...;...)
+    if (processedInput.contains(';') &&
+        processedInput.contains('x') &&
+        processedInput.contains('y')) {
+      return _solveSystemOfLinearEquations(processedInput);
+    }
+
+    // 2. 检查是否为一元二次方程 (包含 x^2 或 x²)
+    if (processedInput.contains('x^2') || processedInput.contains('x²')) {
+      return _solveQuadraticEquation(processedInput.replaceAll('x²', 'x^2'));
+    }
+
+    // 3. 检查是否为一元一次方程 (包含 x 但不包含 y 或 x^2)
+    if (processedInput.contains('x') && !processedInput.contains('y')) {
+      return _solveLinearEquation(processedInput);
+    }
+
+    // 4. 如果都不是，则作为简单表达式计算
+    try {
+      return _solveSimpleExpression(input); // 使用原始输入以保留运算符
+    } catch (e) {
+      if (kDebugMode) {
+        print(e);
+      }
+      throw Exception('无法识别的格式。请检查您的方程或表达式。');
+    }
+  }
+
+  /// ---- 求解器实现 ----
+
+  /// 1. 求解简单表达式
+  CalculationResult _solveSimpleExpression(String input) {
+    final steps = <CalculationStep>[];
+    steps.add(
+      CalculationStep(
+        stepNumber: 1,
+        title: '表达式求值',
+        explanation: '这是一个标准的数学表达式，我们将直接计算其结果。',
+        formula: input,
+      ),
+    );
+
+    GrammarParser p = GrammarParser();
+    Expression exp = p.parse(input);
+    final result = RealEvaluator().evaluate(exp);
+
+    return CalculationResult(steps: steps, finalAnswer: result.toString());
+  }
+
+  /// 2. 求解一元一次方程
+  CalculationResult _solveLinearEquation(String input) {
+    final steps = <CalculationStep>[];
+    steps.add(
+      CalculationStep(
+        stepNumber: 0,
+        title: '原方程',
+        explanation: '这是一元一次方程。',
+        formula: '\$\$$input\$\$',
+      ),
+    );
+
+    final parts = _parseLinearEquation(input);
+    final a = parts.a, b = parts.b, c = parts.c, d = parts.d;
+
+    final newA = a - c;
+    final newD = d - b;
+
+    steps.add(
+      CalculationStep(
+        stepNumber: 1,
+        title: '移项',
+        explanation: '将所有含 x 的项移到等式左边，常数项移到右边。',
+        formula:
+            '\$\$${a}x ${c >= 0 ? '-' : '+'} ${c.abs()}x = $d ${b >= 0 ? '-' : '+'} ${b.abs()}\$\$',
+      ),
+    );
+
+    steps.add(
+      CalculationStep(
+        stepNumber: 2,
+        title: '合并同类项',
+        explanation: '合并等式两边的项。',
+        formula: '\$\$${newA}x = $newD\$\$',
+      ),
+    );
+
+    if (newA == 0) {
+      return CalculationResult(
+        steps: steps,
+        finalAnswer: newD == 0 ? '有无穷多解' : '无解',
+      );
+    }
+
+    final x = newD / newA;
+    steps.add(
+      CalculationStep(
+        stepNumber: 3,
+        title: '求解 x',
+        explanation: '两边同时除以 x 的系数 ($newA)。',
+        formula: '\$\$x = \frac{$newD}{$newA}\$\$',
+      ),
+    );
+
+    return CalculationResult(steps: steps, finalAnswer: '\$\$x = $x\$\$');
+  }
+
+  /// 3. 求解一元二次方程 (升级版)
+  CalculationResult _solveQuadraticEquation(String input) {
+    final steps = <CalculationStep>[];
+
+    final eqParts = input.split('=');
+    if (eqParts.length != 2) throw Exception("方程格式错误，应包含一个 '='。");
+
+    final leftCoeffs = _parsePolynomial(eqParts[0]);
+    final rightCoeffs = _parsePolynomial(eqParts[1]);
+
+    final a = (leftCoeffs[2] ?? 0) - (rightCoeffs[2] ?? 0);
+    final b = (leftCoeffs[1] ?? 0) - (rightCoeffs[1] ?? 0);
+    final c = (leftCoeffs[0] ?? 0) - (rightCoeffs[0] ?? 0);
+
+    if (a == 0) {
+      return _solveLinearEquation('${b}x+$c=0');
+    }
+
+    steps.add(
+      CalculationStep(
+        stepNumber: 1,
+        title: '整理方程',
+        explanation: r'将方程整理成标准形式 ax^2+bx+c=0。',
+        formula:
+            '\$\$${a}x^2 ${b >= 0 ? '+' : ''} ${b}x ${c >= 0 ? '+' : ''} $c = 0\$\$',
+      ),
+    );
+
+    if (a == a.round() && b == b.round() && c == c.round()) {
+      final factors = _tryFactorization(a.toInt(), b.toInt(), c.toInt());
+      if (factors != null) {
+        steps.add(
+          CalculationStep(
+            stepNumber: 2,
+            title: '因式分解法 (十字相乘)',
+            explanation: '我们发现可以将方程分解为两个一次因式的乘积。',
+            formula: factors.formula,
+          ),
+        );
+        steps.add(
+          CalculationStep(
+            stepNumber: 3,
+            title: '求解',
+            explanation: '分别令每个因式等于 0，解出 x。',
+            formula: factors.solution,
+          ),
+        );
+        return CalculationResult(steps: steps, finalAnswer: factors.solution);
+      }
+    }
+
+    steps.add(
+      CalculationStep(
+        stepNumber: 2,
+        title: '选择解法',
+        explanation: '无法进行因式分解，我们选择使用求根公式法。',
+        formula: '\$\$\\Delta = b^2 - 4ac\$\$',
+      ),
+    );
+
+    final delta = b * b - 4 * a * c;
+    steps.add(
+      CalculationStep(
+        stepNumber: 3,
+        title: '计算判别式 (Delta)',
+        explanation:
+            '\$\$\\Delta = b^2 - 4ac = ($b)^2 - 4 \\cdot ($a) \\cdot ($c) = $delta\$\$',
+        formula: '\$\$\\Delta = $delta\$\$',
+      ),
+    );
+
+    if (delta > 0) {
+      final x1 = (-b + sqrt(delta)) / (2 * a);
+      final x2 = (-b - sqrt(delta)) / (2 * a);
+      steps.add(
+        CalculationStep(
+          stepNumber: 4,
+          title: '应用求根公式',
+          explanation:
+              r'因为 $\Delta > 0$，方程有两个不相等的实数根。公式: $x = \frac{-b \pm \sqrt{\Delta}}{2a}$。',
+          formula:
+              '\$\$x_1 = ${x1.toStringAsFixed(4)}, \\quad x_2 = ${x2.toStringAsFixed(4)}\$\$',
+        ),
+      );
+      return CalculationResult(
+        steps: steps,
+        finalAnswer:
+            '\$\$x_1 = ${x1.toStringAsFixed(4)}, \\quad x_2 = ${x2.toStringAsFixed(4)}\$\$',
+      );
+    } else if (delta == 0) {
+      final x = -b / (2 * a);
+      steps.add(
+        CalculationStep(
+          stepNumber: 4,
+          title: '应用求根公式',
+          explanation: r'因为 $\Delta = 0$，方程有两个相等的实数根。',
+          formula: '\$\$x_1 = x_2 = ${x.toStringAsFixed(4)}\$\$',
+        ),
+      );
+      return CalculationResult(
+        steps: steps,
+        finalAnswer: '\$\$x_1 = x_2 = ${x.toStringAsFixed(4)}\$\$',
+      );
+    } else {
+      steps.add(
+        CalculationStep(
+          stepNumber: 4,
+          title: '判断解',
+          explanation: r'因为 $\Delta < 0$，该方程在实数范围内无解。',
+          formula: '无实数解',
+        ),
+      );
+      return CalculationResult(steps: steps, finalAnswer: '无实数解');
+    }
+  }
+
+  /// 4. 求解二元一次方程组
+  CalculationResult _solveSystemOfLinearEquations(String input) {
+    final steps = <CalculationStep>[];
+    final equations = input.split(';');
+    if (equations.length != 2) throw Exception("格式错误, 请用 ';' 分隔两个方程。");
+
+    final p1 = _parseTwoVariableLinear(equations[0]);
+    final p2 = _parseTwoVariableLinear(equations[1]);
+
+    double a1 = p1[0], b1 = p1[1], c1 = p1[2];
+    double a2 = p2[0], b2 = p2[1], c2 = p2[2];
+
+    steps.add(
+      CalculationStep(
+        stepNumber: 0,
+        title: '原始方程组',
+        explanation: '这是一个二元一次方程组，我们将使用加减消元法求解。',
+        formula:
+            '''
+
+\\begin{cases}
+${a1}x ${b1 >= 0 ? '+' : ''} ${b1}y = $c1 & (1) \\
+${a2}x ${b2 >= 0 ? '+' : ''} ${b2}y = $c2 & (2)
+\\end{cases}
+
+''',
+      ),
+    );
+
+    final det = a1 * b2 - a2 * b1;
+    if (det == 0) {
+      return CalculationResult(
+        steps: steps,
+        finalAnswer: a1 * c2 - a2 * c1 == 0 ? '有无穷多解' : '无解',
+      );
+    }
+
+    final newA1 = a1 * b2, newC1 = c1 * b2;
+    final newA2 = a2 * b1, newC2 = c2 * b1;
+
+    steps.add(
+      CalculationStep(
+        stepNumber: 1,
+        title: '消元',
+        explanation: '为了消去变量 y，将方程(1)两边乘以 $b2，方程(2)两边乘以 $b1。',
+        formula:
+            '''
+
+\\begin{cases}
+${newA1}x ${b1 * b2 >= 0 ? '+' : ''} ${b1 * b2}y = $newC1 & (3) \\
+${newA2}x ${b1 * b2 >= 0 ? '+' : ''} ${b1 * b2}y = $newC2 & (4)
+\\end{cases}
+
+''',
+      ),
+    );
+
+    final xCoeff = newA1 - newA2;
+    final constCoeff = newC1 - newC2;
+
+    steps.add(
+      CalculationStep(
+        stepNumber: 2,
+        title: '相减',
+        explanation: '将方程(3)减去方程(4)，得到一个只含 x 的方程。',
+        formula:
+            '\$\$($newA1 - $newA2)x = $newC1 - $newC2 \\Rightarrow ${xCoeff}x = $constCoeff\$\$',
+      ),
+    );
+
+    final x = constCoeff / xCoeff;
+    steps.add(
+      CalculationStep(
+        stepNumber: 3,
+        title: '解出 x',
+        explanation: '求解上述方程得到 x 的值。',
+        formula: '\$\$x = $x\$\$',
+      ),
+    );
+
+    if (b1.abs() < 1e-9) {
+      final yCoeff = b2;
+      final yConst = c2 - a2 * x;
+      final y = yConst / yCoeff;
+      steps.add(
+        CalculationStep(
+          stepNumber: 4,
+          title: '回代求解 y',
+          explanation: '将 x = $x 代入原方程(2)中。',
+          formula:
+              '''
+
+\\begin{aligned}
+$a2($x) + ${b2}y &= $c2 \\
+${a2 * x} + ${b2}y &= $c2 \\
+${b2}y &= $c2 - ${a2 * x} \\
+${b2}y &= ${c2 - a2 * x}
+\\end{aligned}
+
+''',
+        ),
+      );
+      steps.add(
+        CalculationStep(
+          stepNumber: 5,
+          title: '解出 y',
+          explanation: '求解得到 y 的值。',
+          formula: '\$\$y = $y\$\$',
+        ),
+      );
+      return CalculationResult(
+        steps: steps,
+        finalAnswer: '\$\$x = $x, \\quad y = $y\$\$',
+      );
+    } else {
+      final yCoeff = b1;
+      final yConst = c1 - a1 * x;
+      final y = yConst / yCoeff;
+      steps.add(
+        CalculationStep(
+          stepNumber: 4,
+          title: '回代求解 y',
+          explanation: '将 x = $x 代入原方程(1)中。',
+          formula:
+              '''
+
+\\begin{aligned}
+$a1($x) + ${b1}y &= $c1 \\
+${a1 * x} + ${b1}y &= $c1 \\
+${b1}y &= $c1 - ${a1 * x} \\
+${b1}y &= ${c1 - a1 * x}
+\\end{aligned}
+
+''',
+        ),
+      );
+      steps.add(
+        CalculationStep(
+          stepNumber: 5,
+          title: '解出 y',
+          explanation: '求解得到 y 的值。',
+          formula: '\$\$y = $y\$\$',
+        ),
+      );
+      return CalculationResult(
+        steps: steps,
+        finalAnswer: '\$\$x = $x, \\quad y = $y\$\$',
+      );
+    }
+  }
+
+  /// ---- 辅助函数 ----
+
+  String _expandExpressions(String input) {
+    String result = input;
+    while (true) {
+      String oldResult = result;
+
+      final powerMatch = RegExp(
+        r'(-?\d*\.?\d*)?\(([^)]+)\)\^2',
+      ).firstMatch(result);
+      if (powerMatch != null) {
+        final kStr = powerMatch.group(1);
+        double k = 1.0;
+        if (kStr != null && kStr.isNotEmpty) {
+          k = kStr == '-' ? -1.0 : double.parse(kStr);
+        }
+
+        final factor = powerMatch.group(2)!;
+        final coeffs = _parsePolynomial(factor);
+        final a = coeffs[1] ?? 0;
+        final b = coeffs[0] ?? 0;
+
+        final newA = k * a * a;
+        final newB = k * 2 * a * b;
+        final newC = k * b * b;
+
+        final expanded =
+            '${newA}x^2${newB >= 0 ? '+' : ''}${newB}x${newC >= 0 ? '+' : ''}$newC';
+        result = result.replaceFirst(powerMatch.group(0)!, '($expanded)');
+        continue;
+      }
+
+      final factorMulMatch = RegExp(
+        r'\(([^)]+)\)\(([^)]+)\)',
+      ).firstMatch(result);
+      if (factorMulMatch != null) {
+        final factor1 = factorMulMatch.group(1)!;
+        final factor2 = factorMulMatch.group(2)!;
+        final coeffs1 = _parsePolynomial(factor1);
+        final coeffs2 = _parsePolynomial(factor2);
+
+        final a = coeffs1[1] ?? 0;
+        final b = coeffs1[0] ?? 0;
+        final c = coeffs2[1] ?? 0;
+        final d = coeffs2[0] ?? 0;
+
+        final newA = a * c;
+        final newB = a * d + b * c;
+        final newC = b * d;
+
+        final expanded =
+            '${newA}x^2${newB >= 0 ? '+' : ''}${newB}x${newC >= 0 ? '+' : ''}$newC';
+        result = result.replaceFirst(factorMulMatch.group(0)!, '($expanded)');
+        continue;
+      }
+
+      if (result == oldResult) break;
+    }
+    return result;
+  }
+
+  LinearEquationParts _parseLinearEquation(String input) {
+    final parts = input.split('=');
+    if (parts.length != 2) throw Exception("方程格式错误，应包含一个'='。");
+
+    final leftCoeffs = _parsePolynomial(parts[0]);
+    final rightCoeffs = _parsePolynomial(parts[1]);
+
+    return LinearEquationParts(
+      (leftCoeffs[1] ?? 0.0),
+      (leftCoeffs[0] ?? 0.0),
+      (rightCoeffs[1] ?? 0.0),
+      (rightCoeffs[0] ?? 0.0),
+    );
+  }
+
+  Map<int, double> _parsePolynomial(String side) {
+    final coeffs = <int, double>{};
+    final pattern = RegExp(
+      r'([+-]?(?:\d*\.?\d*)?)x(?:\^(\d+))?|([+-]?\d*\.?\d+)',
+    );
+    var s = side.startsWith('+') || side.startsWith('-') ? side : '+$side';
+
+    for (final match in pattern.allMatches(s)) {
+      if (match.group(3) != null) {
+        coeffs[0] = (coeffs[0] ?? 0) + double.parse(match.group(3)!);
+      } else {
+        int power = match.group(2) != null ? int.parse(match.group(2)!) : 1;
+        String coeffStr = match.group(1) ?? '+';
+        double coeff = 1.0;
+        if (coeffStr.isNotEmpty && coeffStr != '+') {
+          coeff = coeffStr == '-' ? -1.0 : double.parse(coeffStr);
+        } else if (coeffStr == '-') {
+          coeff = -1.0;
+        }
+        coeffs[power] = (coeffs[power] ?? 0) + coeff;
+      }
+    }
+    return coeffs;
+  }
+
+  List<double> _parseTwoVariableLinear(String equation) {
+    final parts = equation.split('=');
+    if (parts.length != 2) throw Exception("方程 $equation 格式错误");
+    final c = double.tryParse(parts[1]) ?? 0.0;
+
+    double a = 0, b = 0;
+    final xMatch = RegExp(r'([+-]?\d*\.?\d*)x').firstMatch(parts[0]);
+    if (xMatch != null) {
+      final coeff = xMatch.group(1);
+      if (coeff == null || coeff.isEmpty || coeff == '+') {
+        a = 1.0;
+      } else if (coeff == '-') {
+        a = -1.0;
+      } else {
+        a = double.tryParse(coeff) ?? 0.0;
+      }
+    }
+    final yMatch = RegExp(r'([+-]?\d*\.?\d*)y').firstMatch(parts[0]);
+    if (yMatch != null) {
+      final coeff = yMatch.group(1);
+      if (coeff == null || coeff.isEmpty || coeff == '+') {
+        b = 1.0;
+      } else if (coeff == '-') {
+        b = -1.0;
+      } else {
+        b = double.tryParse(coeff) ?? 0.0;
+      }
+    }
+    return [a, b, c];
+  }
+
+  ({String formula, String solution})? _tryFactorization(int a, int b, int c) {
+    if (a == 0) return null;
+    int ac = a * c;
+    for (int i = 1; i <= sqrt(ac.abs()); i++) {
+      if (ac % i == 0) {
+        int j = ac ~/ i;
+        if (check(i, j, b)) return formatFactor(i, j, a);
+        if (check(-i, -j, b)) return formatFactor(-i, -j, a);
+        if (check(i, -j, b)) return formatFactor(i, -j, a);
+        if (check(-i, j, b)) return formatFactor(-i, j, a);
+      }
+    }
+    return null;
+  }
+
+  bool check(int m, int n, int b) => m + n == b;
+
+  ({String formula, String solution}) formatFactor(int m, int n, int a) {
+    int common = gcd(n.abs(), a.abs());
+    int num = n ~/ common;
+    int den = a ~/ common;
+
+    final a1 = den;
+    final c1 = num;
+    final a2 = a ~/ den;
+    final c2 = m ~/ a2;
+
+    final f1Part1 = a1 == 1 ? 'x' : '${a1}x';
+    final f1 = c1 == 0 ? f1Part1 : '$f1Part1 ${c1 >= 0 ? '+' : ''} $c1';
+
+    final f2Part1 = a2 == 1 ? 'x' : '${a2}x';
+    final f2 = c2 == 0 ? f2Part1 : '$f2Part1 ${c2 >= 0 ? '+' : ''} $c2';
+
+    final int x1Num = -c1, x1Den = a1;
+    final int x2Num = -c2, x2Den = a2;
+
+    final sol1 = x1Den == 1 ? '$x1Num' : '\\frac{$x1Num}{$x1Den}';
+    final sol2 = x2Den == 1 ? '$x2Num' : '\\frac{$x2Num}{$x2Den}';
+
+    final solution = x1Num * x2Den == x2Num * x1Den
+        ? 'x_1 = x_2 = $sol1'
+        : 'x_1 = $sol1, \\quad x_2 = $sol2';
+
+    return (formula: '\$\$($f1)($f2) = 0\$\$', solution: '\$\$$solution\$\$');
+  }
+
+  int gcd(int a, int b) => b == 0 ? a : gcd(b, a % b);
+}