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【单选题】
休克病人可能存在心功能不全的指标是
A.
中心静脉压低，血压低
B.
中心静脉压高，血压低
C.
中心静脉压高，血压正常
D.
中心静脉压低，血压正常
E.
中心静脉压正常，血压高
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参考答案：
参考解析：
刷刷题刷刷变学霸
举一反三
【单选题】下列临床诊断休克的依据，不正确的是（）
A.
有诱发休克的致病因素存在
B.
原有高血压者的收缩压较原水平下降50%以上
C.
意识异常，四肢湿冷，胸骨部位皮肤指压试验阳性（指压后再充盈时间>2秒），黏膜苍白或发绀
D.
脉搏细速>100次／分，尿量<20ml／h或无尿
E.
收缩压<10.7kPa（80mmHg）、脉压<2.67kPa（20mmHg）
【单选题】有如下事件过程：Private Sub Form_Load() Show Dim Arr() As Variant Arr = Array(1, 3, 5, 7, 9, 11, 13, 15) Call converse(Arr) For i = 0 To 7 Print arr(i); Next iEnd Sub 以上程序的功能是，通过调用Converse过程，将数组Arr中的元素逆序存放并输出...
A.
Private Sub converse(dim a() as Variant ) Dim T For i = LBound(A) To UBound(A) T = a(i) a(i) = a(UBound(A) - i) a(UBound(A) - i) = T Next i End Sub
B.
Private Sub converse(dim a() as Variant) Dim T Dim J As Integer J = (LBound(A) + UBound(A)) / 2 For i = LBound(A) To J T = a(i) a(i) = a(UBound(A) - i) a(UBound(A) - i) = T Next i End Sub
C.
Private Sub converse(a() As Variant) Dim T For i = LBound(A) To UBound(A) T = a(i) a(i) = a(UBound(A) - i) a(UBound(A) - i) = T Next i End Sub
D.
Private Sub converse(a() As Variant) Dim T Dim J As Integer J = Int((LBound(A) + UBound(A)) / 2) For i = LBound(A) To J T = a(i) a(i) = a(UBound(A) - i) a(UBound(A) - i) = T Next i End Sub