✨ Support calculate ^0.5

lib/calculator.dart

@@ -1,5 +1,5 @@
 // === 在 abstract class Expr 中添加声明 ===
-import 'dart:math' show sqrt, cos, sin, tan;
+import 'dart:math' show sqrt, cos, sin, tan, pow;
 
 abstract class Expr {
   Expr simplify();
@@ -488,7 +488,7 @@ class SqrtExpr extends Expr {
       SqrtExpr(inner.substitute(varName, value));
 
   @override
-  String toString() => "sqrt($inner)";
+  String toString() => "\\sqrt{${inner.toString()}}";
 }
 
 // === CosExpr ===
@@ -584,6 +584,90 @@ class TanExpr extends Expr {
   String toString() => "tan($inner)";
 }
 
+// === PowExpr ===
+class PowExpr extends Expr {
+  final Expr left, right;
+  PowExpr(this.left, this.right);
+
+  @override
+  Expr simplify() {
+    var l = left.simplify();
+    var r = right.simplify();
+
+    // x^0 = 1
+    if (r is IntExpr && r.value == 0) return IntExpr(1);
+    // x^1 = x
+    if (r is IntExpr && r.value == 1) return l;
+    // 1^x = 1
+    if (l is IntExpr && l.value == 1) return IntExpr(1);
+    // 0^x = 0 (for x != 0)
+    if (l is IntExpr && l.value == 0 && !(r is IntExpr && r.value == 0)) {
+      return IntExpr(0);
+    }
+
+    return PowExpr(l, r);
+  }
+
+  @override
+  Expr evaluate() {
+    var l = left.evaluate();
+    var r = right.evaluate();
+
+    // x^0 = 1
+    if (r is IntExpr && r.value == 0) return IntExpr(1);
+    // x^1 = x
+    if (r is IntExpr && r.value == 1) return l;
+    // 1^x = 1
+    if (l is IntExpr && l.value == 1) return IntExpr(1);
+    // 0^x = 0 (for x != 0)
+    if (l is IntExpr && l.value == 0 && !(r is IntExpr && r.value == 0))
+      return IntExpr(0);
+
+    // If both are numbers, compute
+    if (l is IntExpr && r is IntExpr) {
+      return DoubleExpr(pow(l.value.toDouble(), r.value.toDouble()).toDouble());
+    }
+    if (l is DoubleExpr && r is IntExpr) {
+      return DoubleExpr(pow(l.value, r.value.toDouble()).toDouble());
+    }
+    if (l is IntExpr && r is DoubleExpr) {
+      return DoubleExpr(pow(l.value.toDouble(), r.value).toDouble());
+    }
+    if (l is DoubleExpr && r is DoubleExpr) {
+      return DoubleExpr(pow(l.value, r.value).toDouble());
+    }
+
+    return PowExpr(l, r);
+  }
+
+  @override
+  Expr substitute(String varName, Expr value) => PowExpr(
+    left.substitute(varName, value),
+    right.substitute(varName, value),
+  );
+
+  @override
+  String toString() {
+    String leftStr = left.toString();
+    String rightStr = right.toString();
+
+    // Remove outer parentheses
+    if (leftStr.startsWith('(') && leftStr.endsWith(')')) {
+      leftStr = leftStr.substring(1, leftStr.length - 1);
+    }
+    if (rightStr.startsWith('(') && rightStr.endsWith(')')) {
+      rightStr = rightStr.substring(1, rightStr.length - 1);
+    }
+
+    // Add parentheses around base if it's a complex expression
+    bool needsParens =
+        !(left is VarExpr || left is IntExpr || left is DoubleExpr);
+    String base = needsParens ? '($leftStr)' : leftStr;
+
+    return '$base^{$rightStr}';
+  }
+}
+
 // === 辅助：识别 a * sqrt(X) 形式 ===
 class _SqrtTerm {
   final int coef;

lib/parser.dart

@@ -33,12 +33,12 @@ class Parser {
   }
 
   Expr parseMul() {
-    var expr = parseAtom();
+    var expr = parsePow();
     skipSpaces();
     while (!isEnd && (current == '*' || current == '/')) {
       var op = current;
       eat();
-      var right = parseAtom();
+      var right = parsePow();
       if (op == '*') {
         expr = MulExpr(expr, right);
       } else {
@@ -49,6 +49,17 @@ class Parser {
     return expr;
   }
 
+  Expr parsePow() {
+    var expr = parseAtom();
+    skipSpaces();
+    if (!isEnd && current == '^') {
+      eat();
+      var right = parsePow(); // right associative
+      return PowExpr(expr, right);
+    }
+    return expr;
+  }
+
   Expr parseAtom() {
     skipSpaces();
     if (current == '(') {
@@ -106,13 +117,20 @@ class Parser {
       return VarExpr(varName);
     }
 
-    // 解析整数
+    // 解析数字 (整数或小数)
     var buf = '';
-    while (!isEnd && RegExp(r'\d').hasMatch(current)) {
+    bool hasDot = false;
+    while (!isEnd &&
+        (RegExp(r'\d').hasMatch(current) || (!hasDot && current == '.'))) {
+      if (current == '.') hasDot = true;
       buf += current;
       eat();
     }
     if (buf.isEmpty) throw Exception("无法解析: $current");
+    if (hasDot) {
+      return DoubleExpr(double.parse(buf));
+    } else {
       return IntExpr(int.parse(buf));
     }
   }
+}

lib/solver.dart

@@ -52,12 +52,17 @@ class SolverService {
   /// 1. 求解简单表达式
   CalculationResult _solveSimpleExpression(String input) {
     final steps = <CalculationStep>[];
+    // Parse the input to get LaTeX-formatted version
+    final parser = Parser(input);
+    final parsedExpr = parser.parse();
+    final latexInput = parsedExpr.toString().replaceAll('*', '\\cdot');
+
     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\$\$',
+        formula: '\$\$$latexInput\$\$',
       ),
     );
 
@@ -113,12 +118,17 @@ class SolverService {
   /// 2. 求解一元一次方程
   CalculationResult _solveLinearEquation(String input) {
     final steps = <CalculationStep>[];
+    // Parse the input to get LaTeX-formatted version
+    final parser = Parser(input);
+    final parsedExpr = parser.parse();
+    final latexInput = parsedExpr.toString().replaceAll('*', '\\cdot');
+
     steps.add(
       CalculationStep(
         stepNumber: 0,
         title: '原方程',
         explanation: '这是一元一次方程。',
-        formula: '\$\$$input\$\$',
+        formula: '\$\$$latexInput\$\$',
       ),
     );
 
@@ -1262,8 +1272,7 @@ ${b1}y &= ${c1 - a1 * x.toDouble()}
 
   /// 格式化原始方程，保持符号形式
   String _formatOriginalEquation(String input) {
-    // Simply return the original equation with proper LaTeX formatting
+    // Parse the equation and convert to LaTeX
-    // This avoids complex parsing issues and preserves the original symbolic form
     String result = input.replaceAll(' ', '');
 
     // 确保方程格式正确
@@ -1271,9 +1280,31 @@ ${b1}y &= ${c1 - a1 * x.toDouble()}
       result = '$result=0';
     }
 
-    // Replace sqrt with LaTeX format
+    final parts = result.split('=');
+    if (parts.length == 2) {
+      try {
+        final leftParser = Parser(parts[0]);
+        final leftExpr = leftParser.parse();
+        final rightParser = Parser(parts[1]);
+        final rightExpr = rightParser.parse();
+        result =
+            '${leftExpr.toString().replaceAll('*', '\\cdot')}=${rightExpr.toString().replaceAll('*', '\\cdot')}';
+      } catch (e) {
+        // Fallback to original if parsing fails
         result = result.replaceAll('sqrt(', '\\sqrt{');
         result = result.replaceAll(')', '}');
+      }
+    } else {
+      try {
+        final parser = Parser(result.split('=')[0]);
+        final expr = parser.parse();
+        result = '${expr.toString().replaceAll('*', '\\cdot')}=0';
+      } catch (e) {
+        // Fallback
+        result = result.replaceAll('sqrt(', '\\sqrt{');
+        result = result.replaceAll(')', '}');
+      }
+    }
 
     return '\$\$$result\$\$';
   }

test/calculator_test.dart

@@ -39,7 +39,7 @@ void main() {
 
     test('非完全平方数', () {
       var expr = Parser("sqrt(8)").parse();
-      expect(expr.simplify().toString().replaceAll(' ', ''), "(2*sqrt(2))");
+      expect(expr.simplify().toString().replaceAll(' ', ''), "(2*\\sqrt{2})");
     });
   });
 
@@ -53,7 +53,7 @@ void main() {
       var expr = Parser("sqrt(8)/4 + 1/2").parse();
       expect(
         expr.evaluate().toString().replaceAll(' ', ''),
-        "((sqrt(2)/2)+1/2)",
+        "((\\sqrt{2}/2)+1/2)",
       );
     });
   });