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💄 Optimize solver

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This commit is contained in:
LittleSheep
2025-09-16 19:26:27 +08:00
parent dd4a9f524e
commit 47b6fb853a
1 changed files with 27 additions and 15 deletions
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42
lib/solver.dart
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@@ -12,6 +12,18 @@ class LinearEquationParts {
}
class SolverService {
/// 格式化数字，移除不必要的尾随零
String _formatNumber(double value, {int precision = 4}) {
String formatted = value.toStringAsFixed(precision);
// 移除尾随的零和小数点
formatted = formatted.replaceAll(RegExp(r'\.0+$'), '');
// 如果最后是小数点，也移除
if (formatted.endsWith('.')) {
formatted = formatted.substring(0, formatted.length - 1);
}
return formatted;
}
/// 主入口方法，识别并分发任务
CalculationResult solve(String input) {
// 预处理输入字符串
@@ -248,7 +260,7 @@ class SolverService {
title: '合并同类项',
explanation: '合并等式两边的项。',
formula:
'\$\$${newA.toDouble().toStringAsFixed(4)}x = ${newD.toDouble().toStringAsFixed(4)}\$\$',
'\$\$${_formatNumber(newA.toDouble())}x = ${_formatNumber(newD.toDouble())}\$\$',
),
);
@@ -489,7 +501,7 @@ class SolverService {
if (simplified is IntExpr) {
rootStr = simplified.value.toString();
} else {
rootStr = rootValue.toStringAsFixed(6).replaceAll(RegExp(r'\.0+$'), '');
rootStr = _formatNumber(rootValue.toDouble(), precision: 6);
}
}
@@ -605,7 +617,7 @@ class SolverService {
if (simplified is IntExpr) {
resultStr = simplified.value.toString();
} else {
resultStr = result.toStringAsFixed(6).replaceAll(RegExp(r'\.0+$'), '');
resultStr = _formatNumber(result.toDouble(), precision: 6);
}
}
@@ -680,8 +692,8 @@ ${a2}x ${b2 >= 0 ? '+' : ''} ${b2}y = $c2 & (2)
'''
\$\$
\\begin{cases}
${newA1.toDouble().toStringAsFixed(2)}x ${b1 * b2 >= 0 ? '+' : ''} ${(b1 * b2).toStringAsFixed(2)}y = ${newC1.toDouble().toStringAsFixed(2)} & (3) \\\\
${newA2.toDouble().toStringAsFixed(2)}x ${b1 * b2 >= 0 ? '+' : ''} ${(b1 * b2).toStringAsFixed(2)}y = ${newC2.toDouble().toStringAsFixed(2)} & (4)
${_formatNumber(newA1.toDouble(), precision: 2)}x ${b1 * b2 >= 0 ? '+' : ''} ${_formatNumber((b1 * b2), precision: 2)}y = ${_formatNumber(newC1.toDouble(), precision: 2)} & (3) \\\\
${_formatNumber(newA2.toDouble(), precision: 2)}x ${b1 * b2 >= 0 ? '+' : ''} ${_formatNumber((b1 * b2), precision: 2)}y = ${_formatNumber(newC2.toDouble(), precision: 2)} & (4)
\\end{cases}
\$\$
''',
@@ -697,7 +709,7 @@ ${newA2.toDouble().toStringAsFixed(2)}x ${b1 * b2 >= 0 ? '+' : ''} ${(b1 * b2).t
title: '相减',
explanation: '将方程(3)减去方程(4)，得到一个只含 x 的方程。',
formula:
'\$\$(${newA1.toDouble().toStringAsFixed(2)} - ${newA2.toDouble().toStringAsFixed(2)})x = ${newC1.toDouble().toStringAsFixed(2)} - ${newC2.toDouble().toStringAsFixed(2)} \\Rightarrow ${xCoeff.toDouble().toStringAsFixed(2)}x = ${constCoeff.toDouble().toStringAsFixed(2)}\$\$',
'\$\$(${_formatNumber(newA1.toDouble(), precision: 2)} - ${_formatNumber(newA2.toDouble(), precision: 2)})x = ${_formatNumber(newC1.toDouble(), precision: 2)} - ${_formatNumber(newC2.toDouble(), precision: 2)} \\Rightarrow ${_formatNumber(xCoeff.toDouble(), precision: 2)}x = ${_formatNumber(constCoeff.toDouble(), precision: 2)}\$\$',
),
);
@@ -707,7 +719,7 @@ ${newA2.toDouble().toStringAsFixed(2)}x ${b1 * b2 >= 0 ? '+' : ''} ${(b1 * b2).t
stepNumber: 3,
title: '解出 x',
explanation: '求解上述方程得到 x 的值。',
formula: '\$\$x = $x\$\$',
formula: '\$\$x = ${_formatNumber(x.toDouble())}\$\$',
),
);
@@ -719,12 +731,12 @@ ${newA2.toDouble().toStringAsFixed(2)}x ${b1 * b2 >= 0 ? '+' : ''} ${(b1 * b2).t
CalculationStep(
stepNumber: 4,
title: '回代求解 y',
explanation: '将 x = ${x.toDouble().toStringAsFixed(4)} 代入原方程(2)中。',
explanation: '将 x = ${_formatNumber(x.toDouble())} 代入原方程(2)中。',
formula:
'''
\$\$
\\begin{aligned}
$a2(${x.toDouble().toStringAsFixed(4)}) + ${b2}y &= $c2 \\\\
$a2(${_formatNumber(x.toDouble())}) + ${b2}y &= $c2 \\\\
${a2 * x.toDouble()} + ${b2}y &= $c2 \\\\
${b2}y &= $c2 - ${a2 * x.toDouble()} \\\\
${b2}y &= ${c2 - a2 * x.toDouble()}
@@ -738,13 +750,13 @@ ${b2}y &= ${c2 - a2 * x.toDouble()}
stepNumber: 5,
title: '解出 y',
explanation: '求解得到 y 的值。',
formula: '\$\$y = ${y.toStringAsFixed(4)}\$\$',
formula: '\$\$y = ${_formatNumber(y.toDouble())}\$\$',
),
);
return CalculationResult(
steps: steps,
finalAnswer:
'\$\$x = ${x.toDouble().toStringAsFixed(4)}, \\quad y = ${y.toStringAsFixed(4)}\$\$',
'\$\$x = ${_formatNumber(x.toDouble())}, \\quad y = ${_formatNumber(y.toDouble())}\$\$',
);
} else {
final yCoeff = b1;
@@ -754,12 +766,12 @@ ${b2}y &= ${c2 - a2 * x.toDouble()}
CalculationStep(
stepNumber: 4,
title: '回代求解 y',
explanation: '将 x = ${x.toDouble().toStringAsFixed(4)} 代入原方程(1)中。',
explanation: '将 x = ${_formatNumber(x.toDouble())} 代入原方程(1)中。',
formula:
'''
\$\$
\\begin{aligned}
$a1(${x.toDouble().toStringAsFixed(4)}) + ${b1}y &= $c1 \\\\
$a1(${_formatNumber(x.toDouble())}) + ${b1}y &= $c1 \\\\
${a1 * x.toDouble()} + ${b1}y &= $c1 \\\\
${b1}y &= $c1 - ${a1 * x.toDouble()} \\\\
${b1}y &= ${c1 - a1 * x.toDouble()}
@@ -773,13 +785,13 @@ ${b1}y &= ${c1 - a1 * x.toDouble()}
stepNumber: 5,
title: '解出 y',
explanation: '求解得到 y 的值。',
formula: '\$\$y = ${y.toStringAsFixed(4)}\$\$',
formula: '\$\$y = ${_formatNumber(y.toDouble())}\$\$',
),
);
return CalculationResult(
steps: steps,
finalAnswer:
'\$\$x = ${x.toDouble().toStringAsFixed(4)}, \\quad y = ${y.toStringAsFixed(4)}\$\$',
'\$\$x = ${_formatNumber(x.toDouble())}, \\quad y = ${_formatNumber(y.toDouble())}\$\$',
);
}
}