diff --git a/config.yaml b/config.yaml
index e8cbcac..64eea92 100644
--- a/config.yaml
+++ b/config.yaml
@@ -1,30 +1,20 @@
-# ============================================================
-# KOCR v4 — 配置中心
-# 简化版：原图直出 PaddleOCR，无预处理
-# ============================================================
-
-# OCR 参数（PaddleOCR 原生参数）
 ocr:
-  use_gpu: false
-  lang: ch
-  show_log: false
-  use_angle_cls: true
-  det_db_thresh: 0.3
   det_db_box_thresh: 0.5
-  det_db_unclip_ratio: 1.5
-  max_side_len: 1600
   det_db_score_mode: fast
+  det_db_thresh: 0.3
+  det_db_unclip_ratio: 1.5
+  lang: ch
+  max_side_len: 1600
   min_size: 8
   rec_drop_score: 0.5
-
-# 后处理参数
-postprocessing:
-  amount_balance_tolerance: 0.01
-
-# 输出路径
+  show_log: false
+  use_angle_cls: true
+  use_gpu: true
 paths:
-  input_dir: input
-  output_dir: output
   done_dir: input/done
   index_file: .kocr_index.json
+  input_dir: input
+  output_dir: output
   template: /home/muc/mc/会计工具/kocr-v4/凭证模板.xls
+postprocessing:
+  amount_balance_tolerance: 0.01
diff --git a/core/parser.py b/core/parser.py
index 2de5183..318a08e 100644
--- a/core/parser.py
+++ b/core/parser.py
@@ -374,59 +374,39 @@ class K3VoucherParser:
         return ''
     
     def _match_entries(self, result, codes, debit_amts, credit_amts, summary, bounds):
-        """将金额按匹配到科目，构建分录"""
+        """将金额匹配到科目，构建分录
+        
+        顺序分配 + 多余金额合并到最后一个科目：
+        - 第1个借方金额 → 第1个科目，第2个借方 → 第2个科目……
+        - 借方金额多于科目数 → 多余金额合并到最后一个科目
+        - 贷方同理
+        - 不做平均分配（平均分配会产生不存在于原始凭证的金额）
+        """
         if not codes:
             result['entries'] = []
             return
         
         n_codes = len(codes)
-        n_debit = len(debit_amts)
-        n_credit = len(credit_amts)
         
-        # 策略 1：等额借贷
-        if n_codes == n_debit == n_credit:
-            for i, code in enumerate(codes):
-                code['debit'] = debit_amts[i]['val']
-                code['credit'] = credit_amts[i]['val']
+        # 初始化
+        for code in codes:
+            code['debit'] = 0.0
+            code['credit'] = 0.0
         
-        # 策略 2：单行分录
-        elif n_codes == 1 and (n_debit > 0 or n_credit > 0):
-            codes[0]['debit'] = sum(a['val'] for a in debit_amts)
-            codes[0]['credit'] = sum(a['val'] for a in credit_amts)
+        # 借方顺序分配：多余的累加到最后一个科目
+        for i, amt in enumerate(debit_amts):
+            idx = min(i, n_codes - 1)
+            codes[idx]['debit'] += amt['val']
         
-        # 策略 3：借贷均足量，按顺序匹配
-        elif n_debit >= n_codes and n_credit >= n_codes:
-            for i, code in enumerate(codes):
-                if i < n_debit:
-                    code['debit'] = debit_amts[i]['val']
-                if i < n_credit:
-                    code['credit'] = credit_amts[i]['val']
+        # 贷方顺序分配：多余的累加到最后一个科目
+        for i, amt in enumerate(credit_amts):
+            idx = min(i, n_codes - 1)
+            codes[idx]['credit'] += amt['val']
         
-        # 策略 4：借方多，贷方平均分配
-        elif n_debit >= n_codes:
-            for i, code in enumerate(codes):
-                code['debit'] = debit_amts[i]['val']
-            if n_credit > 0:
-                per_code = sum(a['val'] for a in credit_amts) / n_codes
-                for code in codes:
-                    code['credit'] = round(per_code, 2)
-        
-        # 策略 5：贷方多，借方平均分配
-        elif n_credit >= n_codes:
-            for i, code in enumerate(codes):
-                code['credit'] = credit_amts[i]['val']
-            if n_debit > 0:
-                per_code = sum(a['val'] for a in debit_amts) / n_codes
-                for code in codes:
-                    code['debit'] = round(per_code, 2)
-        
-        # 策略 6：都不够
-        else:
-            for i, code in enumerate(codes):
-                if i < n_debit:
-                    code['debit'] = debit_amts[i]['val']
-                if i < n_credit:
-                    code['credit'] = credit_amts[i]['val']
+        # 四舍五入
+        for code in codes:
+            code['debit'] = round(code['debit'], 2)
+            code['credit'] = round(code['credit'], 2)
         
         # 设置摘要
         for code in codes: