如果结果不匹配，请 
更多“配制除菌GH时蔗糖的终浓度为（)。”相关的问题
A.伤口深处的细菌在潮湿的环境中生长旺盛，而蔗糖有脱水的作用。
B.许多种细菌以蔗糖为养料，当它们得到糖时迅速繁殖。
C.有些含蔗糖的食物能够削弱某些抗菌素的作用。
D.高效的抗菌素只是最近才开发出来的，用蔗糖治疗伤口则有久远的历史。
PB I/E CORPORATION
BUSAN KOREA OUR BOOK No.:
24JFK5466J B/L No.:
ZW780321 CONSIGNEE:
HENAN GH IMP/EXP CO., LTD．
No.34 GUANGZHOU ROAD, ZHENGZHOU, CHINA REMARKS:
NOTIFY PARTY:
DALIAN DAXING COMMERCIAL&TRADE CO., LTD．
No.176 ZHONGSHAN ROAD, DALIAN, CHINA PORT OF LOADING：
SYDNEY VESSEL:
STAR RIVER
VOYAGE No.:
847E
FLAG:
GERMANY PORT OF DISCHARGE:
DALIAN CHINA VIA INCHON PLACE OF DELIVERY: MARK
No.OF PKGS
DESCRIPTION OF GOODS
GROSS WEIGHT
MEASUREMENT
GH 20 PLASTIC PALLETS 18,000KGS 17,600CBM
生羊皮的检验检疫监管类别是M.P/N.Q，说明该种货物进境时需实施品质检验和动植物检疫。（）
PB I/E CORPORATION
BUSAN KOREA OUR BOOK No.:
24JFK5466J B/L No.:
ZW780321 CONSIGNEE:
HENAN GH IMP/EXP CO., LTD．
No.34 GUANGZHOU ROAD, ZHENGZHOU, CHINA REMARKS:
NOTIFY PARTY:
DALIAN DAXING COMMERCIAL&TRADE CO., LTD．
No.176 ZHONGSHAN ROAD, DALIAN, CHINA PORT OF LOADING：
SYDNEY VESSEL:
STAR RIVER
VOYAGE No.:
847E
FLAG:
GERMANY PORT OF DISCHARGE:
DALIAN CHINA VIA INCHON PLACE OF DELIVERY: MARK
No.OF PKGS
DESCRIPTION OF GOODS
GROSS WEIGHT
MEASUREMENT
GH 20 PLASTIC PALLETS 18,000KGS 17,600CBM
报检时须提供“3C”证书。（）
CONSIGNOR:
PB I/E CORPORATION
BUSAN KOREA
OUR BOOK No.:
24JFK5466J
B/L No.:
ZW780321
CONSIGNEE:
HENAN GH IMP/EXP CO., LTD.
No.34 GUANGZHOU ROAD, ZHENGZHOU, CHINA
REMARKS:
NOTIFY PARTY:
DALIAN DAXING COMMERCIAL&TRADE CO., LTD.
No.176 ZHONGSHAN ROAD, DALIAN, CHINA
PORT OF LOADING：
SYDNEY
VESSEL:
STAR RIVER
VOYAGE No.:
847E
FLAG:
GERMANY
PORT OF DISCHARGE:
DALIAN CHINA VIA INCHON
PLACE OF DELIVERY:
MARK
No.OF PKGS
DESCRIPTION OF GOODS
GROSS WEIGHT
MEASUREMENT
GH 20 PLASTIC PALLETS 18,000KGS 17,600CBM
SHEEP SKINS (AUSTRALIA ORIGIN)
2000PIECES PACKING: IN PLASTIC PALLETS
CONTRACT No.: GH-002
1×20’ CONTAINER
No. MBLU4459040/771126
DATE: Jun. 23, 2004 LUCKY FERRY CO., LTD.
BY_________________________________ BY__________________________________
第131题：生羊皮的检验检疫监管类别是M.P/N.Q，说明该种货物进境时需实施品质检验和动植物检疫。（）
此题为判断题(对，错)。
（31）有以下程序
#include ＜stdio.h＞
main（）
{ char *a[ ]={“abcd”,”ef”,”gh”,”ijk”};int I;
for （i=0;i＜4;i++） printf（“%c”,*a）;
}
程序运行后输出的结果是
A）aegi B）dfhk C）abcd D）abcdefghijk
