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【单选题】
男性 35 岁,右上腹空腹痛反复发作 5 年,近一段加重,近三天黑便(成形),为进一步明确诊断 选检查
A.
胃液分析
B.
大便潜血
C.
胃肠道钡餐
D.
胃镜
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参考答案:
举一反三
【多选题】轻度中暑的临床表现有( )。
A.
面色潮红
B.
血压下降
C.
肌肉痉挛
D.
昏迷
E.
晕厥
【多选题】以下适合用无领导小组讨论进行人员选拔的岗位是( )。
A.
人力资源部主管
B.
技术开发人员
C.
销售部门经理
D.
公关部门经理
E.
副总经理
F.
服务员
【单选题】At which point of the PPF curve is efficient?
A.
inside the curve
B.
on the curve
C.
outside the curve
D.
behind the curve
【单选题】The market supply curve is:
A.
downward sloping and is flatter than an individual's supply curve.
B.
upward sloping and is flatter than an individual's supply curve.
C.
downward sloping and is steeper than an individual's supply curve.
D.
upward sloping and is steeper than an individual's supply curve.
【简答题】2 If you want to use a sharing bike, you must first _________.A. download an APP and pay for the trip B. have a smart phone and download an APP C. unlock the bike and download an APP D. find a nearest...
【单选题】以下不适合用无领导小组讨论进行人员选拔的岗位是( )。
A.
人力资源主管
B.
技术研发人员
C.
销售部门经理
D.
公关部门经理
【单选题】The demand curve of a monopolist is:
A.
is identical to the marginal cost curve.
B.
downward sloping and above the marginal revenue curve.
C.
downward sloping and below the marginal revenue curve.
D.
kinked because of recognized interdependence with other firms.
【单选题】以下与库函数strcpy(char*p1,char*p2)功能不等的程序段是( )。
A.
strcpy1(char *p1,char *p2) {while ((*p1++=*p2++)!='\0');}
B.
strcpy2(char *p1,char *p2) {while ((*p1=*p2)!='\0'){p1++;p2++;}}
C.
strcpy3(char *p1,char *p2) {while (*p1++=*p2++);}
D.
strcpy4(char *p1,char *p2) {while (*p2) *p1++=*p2++;}
【单选题】If you want to use a sharing bike, you must first _________.
A.
download an APP and pay for the trip
B.
have a smart phone and download an APP
C.
unlock the bike and download an APP
D.
find a nearest bike and borrow it from anyone
【单选题】以下与库函数strcpy(char *p1,char *p2)功能不相等的程序段是( )。
A.
void strcpy4(char *p1,char *p2) { while (*p2)  *p1++=*p2++ ; }
B.
void strcpy1(char *p1,char *p2) { while ((*p1++=*p2++)!='\0') ; }
C.
void strcpy2(char *p1,char *p2) { while ((*p1=*p2)!='\0') { p1++; p2++; } }
D.
void strcpy3(char *p1,char *p2) { while (*p1++=*p2++) ; }
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