fix: avoid GetOutputShape CGO crash by using fixed output dimensions

internal/captcha/handler.go

@@ -311,13 +311,16 @@ func (h *Handler) Math(imageBase64 string) (result string, err error) {
 		return "", err
 	}
 
-	// 获取输出形状
+	// 不调用 GetOutputShape，直接从 output 大小推断
-	outputShape, err := sess.GetOutputShape(0)
+	// Math 模型输出固定为 [51, 1, 19] = 969 个元素 或 [19, 1, 51] = 969 个元素
-	if err != nil {
+	outputLen := len(output)
-		return "", fmt.Errorf("获取输出形状失败: %v", err)
+	if outputLen == 0 {
+		return "", fmt.Errorf("模型输出为空")
 	}
 
-	expr := decodeMath(output, outputShape)
+	// 推断输出形状：已知 numChars=51, batch=1, timesteps=19, total=969
+	// 尝试两种格式
+	expr := decodeMathFromOutput(output)
 	if expr == "" {
 		return "", fmt.Errorf("无法识别表达式")
 	}
@@ -388,6 +391,95 @@ func preprocessMath(img image.Image) ([]float32, error) {
 	return pixels, nil
 }
 
+func decodeMathFromOutput(output []float32) string {
+	// Math 模型输出固定: numChars=51, batch=1, timesteps=19, total=969
+	// 可能的格式: [51, 1, 19] 或 [19, 1, 51]
+	numChars := 51
+	batch := 1
+	timesteps := 19
+	totalElems := numChars * batch * timesteps // 969
+	
+	if len(output) != totalElems {
+		// 尝试推断 timesteps
+		timesteps = len(output) / numChars
+		if timesteps <= 0 {
+			return ""
+		}
+	}
+	
+	// 尝试两种格式
+	// 格式1: [T, B, C] = [19, 1, 51]
+	result1 := decodeMathFormat(output, timesteps, batch, numChars, true)
+	// 格式2: [C, B, T] = [51, 1, 19]
+	result2 := decodeMathFormat(output, timesteps, batch, numChars, false)
+	
+	// 返回更长的结果（更可能是正确的）
+	if len(result1) >= len(result2) {
+		return result1
+	}
+	return result2
+}
+
+func decodeMathFormat(output []float32, timesteps, batch, numChars int, isTBC bool) string {
+	var data []float32
+	
+	if isTBC {
+		// 格式: [T, B, C] -> [B, T, C]
+		data = make([]float32, batch*timesteps*numChars)
+		for t := 0; t < timesteps; t++ {
+			for b := 0; b < batch; b++ {
+				for c := 0; c < numChars; c++ {
+					srcIdx := t*batch*numChars + b*numChars + c
+					dstIdx := b*timesteps*numChars + t*numChars + c
+					if srcIdx < len(output) && dstIdx < len(data) {
+						data[dstIdx] = output[srcIdx]
+					}
+				}
+			}
+		}
+	} else {
+		// 格式: [C, B, T] -> [B, T, C]
+		data = make([]float32, batch*timesteps*numChars)
+		for c := 0; c < numChars; c++ {
+			for b := 0; b < batch; b++ {
+				for t := 0; t < timesteps; t++ {
+					srcIdx := c*batch*timesteps + b*timesteps + t
+					dstIdx := b*timesteps*numChars + t*numChars + c
+					if srcIdx < len(output) && dstIdx < len(data) {
+						data[dstIdx] = output[srcIdx]
+					}
+				}
+			}
+		}
+	}
+	
+	// CTC 解码
+	result := ""
+	lastIdx := -1
+
+	for t := 0; t < timesteps; t++ {
+		maxIdx := 0
+		maxProb := float32(-math.MaxFloat32)
+
+		for c := 0; c < numChars; c++ {
+			idx := t*numChars + c
+			if idx < len(data) && data[idx] > maxProb {
+				maxProb = data[idx]
+				maxIdx = c
+			}
+		}
+
+		if maxIdx != 0 && maxIdx != lastIdx {
+			if maxIdx-1 < len(mathChars) {
+				result += string(mathChars[maxIdx-1])
+			}
+		}
+		lastIdx = maxIdx
+	}
+
+	return result
+}
+
 func decodeMath(output []float32, outputShape []int64) string {
 	// 原版逻辑：
 	// if preds.shape[1] == 1: # Batch size is 1 at dim 1 implies [T, B, C]