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\$\$',
      ),
    );

    // 检查是否为特殊三角函数值，可以返回精确结果
    final exactTrigResult = _getExactTrigResult(input);
    if (exactTrigResult != null) {
      return CalculationResult(
        steps: steps,
        finalAnswer: '\$\$$exactTrigResult\$\$',
      );
    }

    // 预处理输入，将三角函数的参数从度转换为弧度
    String processedInput = _convertTrigToRadians(input);

    GrammarParser p = GrammarParser();
    Expression exp = p.parse(processedInput);
    final result = RealEvaluator().evaluate(exp).toDouble();

    // 尝试将结果格式化为几倍根号的形式
    final formattedResult = _formatSqrtResult(result);

    return CalculationResult(
      steps: steps,
      finalAnswer: '\$\$$formattedResult\$\$',
    );
  }

  /// 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,
          ),
        );
        steps.add(
          CalculationStep(
            stepNumber: 4,
            title: '化简结果',
            explanation: '将分数化简到最简形式，并将负号写在分数外面。',
            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，但有虚数解。',
          formula: '无实数解，有虚数解',
        ),
      );

      final sqrtDelta = sqrt(-delta);
      final realPart = -b / (2 * a);
      final imagPart = sqrtDelta / (2 * a);

      steps.add(
        CalculationStep(
          stepNumber: 5,
          title: '计算虚数根',
          explanation: '使用求根公式计算虚数根。',
          formula: r'$$x = \frac{-b \pm \sqrt{-\Delta} i}{2a}$$',
        ),
      );

      return CalculationResult(
        steps: steps,
        finalAnswer:
            '\$\$x_1 = ${realPart.toStringAsFixed(4)} + ${imagPart.toStringAsFixed(4)}i, \\quad x_2 = ${realPart.toStringAsFixed(4)} - ${imagPart.toStringAsFixed(4)}i\$\$',
      );
    }
  }

  /// 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 ----

  /// 获取精确三角函数结果
  String? _getExactTrigResult(String input) {
    final cleanInput = input.replaceAll(' ', '').toLowerCase();

    // 匹配 sin(角度) 模式
    final sinMatch = RegExp(r'^sin\((\d+(?:\+\d+)*)\)$').firstMatch(cleanInput);
    if (sinMatch != null) {
      final angleExpr = sinMatch.group(1)!;
      final angle = _evaluateAngleExpression(angleExpr);
      if (angle != null) {
        return _getSinExactValue(angle);
      }
    }

    // 匹配 cos(角度) 模式
    final cosMatch = RegExp(r'^cos\((\d+(?:\+\d+)*)\)$').firstMatch(cleanInput);
    if (cosMatch != null) {
      final angleExpr = cosMatch.group(1)!;
      final angle = _evaluateAngleExpression(angleExpr);
      if (angle != null) {
        return _getCosExactValue(angle);
      }
    }

    // 匹配 tan(角度) 模式
    final tanMatch = RegExp(r'^tan\((\d+(?:\+\d+)*)\)$').firstMatch(cleanInput);
    if (tanMatch != null) {
      final angleExpr = tanMatch.group(1)!;
      final angle = _evaluateAngleExpression(angleExpr);
      if (angle != null) {
        return _getTanExactValue(angle);
      }
    }

    return null;
  }

  /// 计算角度表达式（如 30+45 = 75）
  int? _evaluateAngleExpression(String expr) {
    final parts = expr.split('+');
    int sum = 0;
    for (final part in parts) {
      final num = int.tryParse(part.trim());
      if (num == null) return null;
      sum += num;
    }
    return sum;
  }

  /// 获取 sin 的精确值
  String? _getSinExactValue(int angle) {
    // 标准化角度到 0-360 度
    final normalizedAngle = angle % 360;

    switch (normalizedAngle) {
      case 0:
      case 360:
        return '0';
      case 30:
        return '\\frac{1}{2}';
      case 45:
        return '\\frac{\\sqrt{2}}{2}';
      case 60:
        return '\\frac{\\sqrt{3}}{2}';
      case 75:
        return '1 + \\frac{\\sqrt{2}}{2}';
      case 90:
        return '1';
      case 120:
        return '\\frac{\\sqrt{3}}{2}';
      case 135:
        return '\\frac{\\sqrt{2}}{2}';
      case 150:
        return '\\frac{1}{2}';
      case 180:
        return '0';
      case 210:
        return '-\\frac{1}{2}';
      case 225:
        return '-\\frac{\\sqrt{2}}{2}';
      case 240:
        return '-\\frac{\\sqrt{3}}{2}';
      case 270:
        return '-1';
      case 300:
        return '-\\frac{\\sqrt{3}}{2}';
      case 315:
        return '-\\frac{\\sqrt{2}}{2}';
      case 330:
        return '-\\frac{1}{2}';
      default:
        return null;
    }
  }

  /// 获取 cos 的精确值
  String? _getCosExactValue(int angle) {
    // cos(angle) = sin(90 - angle)
    final complementaryAngle = 90 - angle;
    return _getSinExactValue(complementaryAngle.abs());
  }

  /// 获取 tan 的精确值
  String? _getTanExactValue(int angle) {
    // tan(angle) = sin(angle) / cos(angle)
    final sinValue = _getSinExactValue(angle);
    final cosValue = _getCosExactValue(angle);

    if (sinValue != null && cosValue != null) {
      if (cosValue == '0') return null; // 未定义
      return '\\frac{$sinValue}{$cosValue}';
    }

    return null;
  }

  /// 将三角函数的参数从度转换为弧度
  String _convertTrigToRadians(String input) {
    String result = input;

    // 正则表达式匹配三角函数调用，如 sin(30), cos(45), tan(60)
    final trigPattern = RegExp(
      r'(sin|cos|tan|asin|acos|atan)\s*\(\s*([^)]+)\s*\)',
      caseSensitive: false,
    );

    result = result.replaceAllMapped(trigPattern, (match) {
      final func = match.group(1)!;
      final arg = match.group(2)!;

      // 如果参数已经是弧度相关的表达式（包含 pi 或 π），则不转换
      if (arg.contains('pi') || arg.contains('π') || arg.contains('rad')) {
        return '$func($arg)';
      }

      // 将度数转换为弧度：度 * π / 180
      return '$func(($arg)*($pi/180))';
    });

    return result;
  }

  /// 将数值结果格式化为几倍根号的形式
  String _formatSqrtResult(double result) {
    // 处理负数
    if (result < 0) {
      return '-${_formatSqrtResult(-result)}';
    }

    // 处理零
    if (result == 0) return '0';

    // 检查是否接近整数
    final rounded = result.round();
    if ((result - rounded).abs() < 1e-10) {
      return rounded.toString();
    }

    // 计算 result 的平方，看它是否接近整数
    final squared = result * result;
    final squaredRounded = squared.round();

    // 如果 squared 接近整数，说明 result 是某个数的平方根
    if ((squared - squaredRounded).abs() < 1e-6) {
      // 寻找最大的完全平方数因子
      int maxSquareFactor = 1;
      for (int i = 2; i * i <= squaredRounded; i++) {
        if (squaredRounded % (i * i) == 0) {
          maxSquareFactor = i * i;
        }
      }

      final coefficient = sqrt(maxSquareFactor).round();
      final remaining = squaredRounded ~/ maxSquareFactor;

      if (remaining == 1) {
        // 完全平方数，直接返回系数
        return coefficient.toString();
      } else if (coefficient == 1) {
        return '\\sqrt{$remaining}';
      } else {
        return '$coefficient\\sqrt{$remaining}';
      }
    }

    // 如果不是平方根的结果，返回原始数值（保留几位小数）
    return result
        .toStringAsFixed(6)
        .replaceAll(RegExp(r'\.0+$'), '')
        .replaceAll(RegExp(r'\.$'), '');
  }

  String _expandExpressions(String input) {
    String result = input;
    int maxIterations = 10; // Prevent infinite loops
    int iterationCount = 0;

    while (iterationCount < maxIterations) {
      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)');
        iterationCount++;
        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)');
        iterationCount++;
        continue;
      }

      // Handle expressions like x(expr) or (expr)x or coeff(expr)
      final termFactorMatch = RegExp(
        r'([+-]?(?:\d*\.?\d*)?x?)\(([^)]+)\)',
      ).firstMatch(result);
      if (termFactorMatch != null) {
        final termStr = termFactorMatch.group(1)!;
        final factorStr = termFactorMatch.group(2)!;

        // Skip if the term is just a sign or empty
        if (termStr == '+' || termStr == '-' || termStr.isEmpty) {
          break;
        }

        // Parse the term (coefficient and x power)
        final termCoeffs = _parsePolynomial(termStr);
        final factorCoeffs = _parsePolynomial(factorStr);

        final termA = termCoeffs[1] ?? 0; // x coefficient
        final termB = termCoeffs[0] ?? 0; // constant term

        final factorA = factorCoeffs[1] ?? 0; // x coefficient
        final factorB = factorCoeffs[0] ?? 0; // constant term

        // Multiply: (termA*x + termB) * (factorA*x + factorB)
        final newA = termA * factorA;
        final newB = termA * factorB + termB * factorA;
        final newC = termB * factorB;

        final expanded =
            '${newA}x^2${newB >= 0 ? '+' : ''}${newB}x${newC >= 0 ? '+' : ''}$newC';
        result = result.replaceFirst(termFactorMatch.group(0)!, '($expanded)');
        iterationCount++;
        continue;
      }

      if (result == oldResult) break;
      iterationCount++;
    }

    if (iterationCount >= maxIterations) {
      throw Exception('表达式展开过于复杂，请简化输入。');
    }

    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>{};
    // 扩展模式以支持 sqrt 函数
    final pattern = RegExp(
      r'([+-]?(?:\d*\.?\d*|sqrt\(\d+\)))x(?:\^(\d+))?|([+-]?(?:\d*\.?\d*|sqrt\(\d+\)))',
    );
    var s = side.startsWith('+') || side.startsWith('-') ? side : '+$side';

    for (final match in pattern.allMatches(s)) {
      if (match.group(3) != null) {
        // 常数项
        final constStr = match.group(3)!;
        final constValue = _parseCoefficientWithSqrt(constStr);
        coeffs[0] = (coeffs[0] ?? 0) + constValue;
      } else {
        // x 的幂次项
        int power = match.group(2) != null ? int.parse(match.group(2)!) : 1;
        String coeffStr = match.group(1) ?? '+';
        final coeff = _parseCoefficientWithSqrt(coeffStr);
        coeffs[power] = (coeffs[power] ?? 0) + coeff;
      }
    }
    return coeffs;
  }

  /// 解析包含 sqrt 函数的系数
  double _parseCoefficientWithSqrt(String coeffStr) {
    if (coeffStr.isEmpty || coeffStr == '+') return 1.0;
    if (coeffStr == '-') return -1.0;

    // 检查是否包含 sqrt 函数
    final sqrtMatch = RegExp(r'sqrt\((\d+)\)').firstMatch(coeffStr);
    if (sqrtMatch != null) {
      final innerValue = int.parse(sqrtMatch.group(1)!);

      // 对于完全平方数，直接返回整数结果
      final sqrtValue = sqrt(innerValue.toDouble());
      final rounded = sqrtValue.round();
      if ((sqrtValue - rounded).abs() < 1e-10) {
        // 检查是否有系数
        final coeffPart = coeffStr.replaceFirst(sqrtMatch.group(0)!, '');
        if (coeffPart.isEmpty) return rounded.toDouble();
        if (coeffPart == '-') return -rounded.toDouble();

        final coeff = double.parse(coeffPart);
        return coeff * rounded;
      }

      // 对于非完全平方数，计算数值但保持高精度
      final nonPerfectSqrtValue = sqrt(innerValue.toDouble());

      // 检查是否有系数
      final coeffPart = coeffStr.replaceFirst(sqrtMatch.group(0)!, '');
      if (coeffPart.isEmpty) return nonPerfectSqrtValue;
      if (coeffPart == '-') return -nonPerfectSqrtValue;

      final coeff = double.parse(coeffPart);
      return coeff * nonPerfectSqrtValue;
    }

    // 普通数值
    return double.parse(coeffStr);
  }

  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;
    int absAc = ac.abs();
    for (int d = 1; d <= sqrt(absAc).toInt(); d++) {
      if (absAc % d == 0) {
        int d1 = d;
        int d2 = absAc ~/ d;
        List<int> signs1 = ac >= 0 ? [1, -1] : [1, -1];
        List<int> signs2 = ac >= 0 ? [1, -1] : [1, -1];
        for (int s1 in signs1) {
          for (int s2 in signs2) {
            int m = s1 * d1;
            int n = s2 * d2;
            if (check(m, n, b)) return formatFactor(m, n, a);
            m = s1 * d1;
            n = s2 * (-d2);
            if (check(m, n, b)) return formatFactor(m, n, a);
            m = s1 * (-d1);
            n = s2 * d2;
            if (check(m, n, b)) return formatFactor(m, n, a);
            m = s1 * (-d1);
            n = s2 * (-d2);
            if (check(m, n, b)) return formatFactor(m, n, 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) {
    // Roots are -m/a and -n/a
    int g1 = gcd(m.abs(), a.abs());
    int root1Num = -m ~/ g1;
    int root1Den = a ~/ g1;

    int g2 = gcd(n.abs(), a.abs());
    int root2Num = -n ~/ g2;
    int root2Den = a ~/ g2;

    String sol1 = _formatFraction(root1Num, root1Den);
    String sol2 = _formatFraction(root2Num, root2Den);

    // For formula, show (a x + m)(x + n/a) or simplified
    String f1 = a == 1 ? 'x' : '${a}x';
    f1 = m == 0 ? f1 : '$f1 ${m >= 0 ? '+' : ''} $m';

    String f2;
    if (n % a == 0) {
      int coeff = n ~/ a;
      f2 = 'x ${coeff >= 0 ? '+' : ''} $coeff';
      if (coeff == 0) f2 = 'x';
    } else {
      f2 = 'x ${n >= 0 ? '+' : ''} \\frac{$n}{$a}';
    }

    String formula = '\$\$($f1)($f2) = 0\$\$';

    String solution;
    if (root1Num * root2Den == root2Num * root1Den) {
      solution = '\$\$x_1 = x_2 = $sol1\$\$';
    } else {
      solution = '\$\$x_1 = $sol1, \\quad x_2 = $sol2\$\$';
    }

    return (formula: formula, solution: solution);
  }

  String _formatFraction(int num, int den) {
    if (den == 0) return 'undefined';

    // Handle sign: make numerator positive, put sign outside
    bool isNegative = (num < 0) != (den < 0);
    int absNum = num.abs();
    int absDen = den.abs();

    // Simplify fraction
    int g = gcd(absNum, absDen);
    absNum ~/= g;
    absDen ~/= g;

    if (absDen == 1) {
      return isNegative ? '-$absNum' : '$absNum';
    } else {
      String fraction = '\\frac{$absNum}{$absDen}';
      return isNegative ? '-$fraction' : fraction;
    }
  }

  int gcd(int a, int b) => b == 0 ? a : gcd(b, a % b);
}