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);
}