if (i<1) then continue else a=1 end if

这段代码是错误的，因为在 Fortran 中，`if` 语句必须包含 `then` 分支和 `else` 分支，不能省略其中的任何一个。同时，`continue` 语句只能在循环语句中使用，不能在条件语句中使用。

如果你想要实现的功能是：当 `i` 小于 1 时跳过循环，否则执行 `a=1`，可以这样写：

do i = 1, n
    if (i < 1) then
        cycle
    else
        a = 1
    endif
    ! 其他处理语句
enddo

上述代码中，当 `i` 小于 1 时，`cycle` 语句会跳过当前循环中的剩余语句，直接进入下一个循环，相当于实现了跳过循环的功能。否则，执行 `a=1`，继续处理其他语句。

相关问题

Dim arrStim(140), Typecount(3),arrtmp(28) As Integer Dim i, k, flag As Integer Dim bools As Boolean For i = 1 To UBound(arrtmp) If i<21 Then k=1 arrtmp(i)=k Else If i<25 Then k=2 arrtmp(i)=k Else k=3 arrtmp(i)=k End If End If Next For i=1 To 5 Do bools=True RandomizeArray arrtmp,1,UBound(arrtmp) If i<>1 And i<>5 Then If arrtmp(1)=2 Or arrtmp(1)=3 Then If arrtmp(1)=flag Then bools=False Else For k=1 To 27 If arrtmp(k)=2 Or arrtmp(k)=3 Then If arrtmp(k)=arrtmp(k+1) Then bools=False Exit For End If End If Next End If Else For k=1 To 27 If arrtmp(k)=2 Or arrtmp(k)=3 Then If arrtmp(k)=arrtmp(k+1) Then bools=False Exit For End If End If Next End If ElseIf i=1 Then If arrtmp(1)=2 Or arrtmp(1)=3 Or arrtmp(2)=2 Or arrtmp(2)=3 Then bools=False Else For k=1 To 27 If arrtmp(k)=2 Or arrtmp(k)=3 Then If arrtmp(k)=arrtmp(k+1) Then bools=False Exit For End If End If Next End If ElseIf i=5 Then If arrtmp(27)=2 Or arrtmp(27)=3 Or arrtmp(28)=2 Or arrtmp(28)=3 Then bools=False ElseIf arrtmp(1)=2 Or arrtmp(1)=3 Then If arrtmp(1)=flag Then bools=False Else For k=1 To 27 If arrtmp(k)=2 Or arrtmp(k)=3 Then If arrtmp(k)=arrtmp(k+1) Then bools=False Exit For End If End If Next End If Else For k=1 To 27 If arrtmp(k)=2 Or arrtmp(k)=3 Then If arrtmp(k)=arrtmp(k+1) Then bools=False Exit For End If End If Next End If End If Loop Until bools=True For k=1 To 28 arrStim((i-1)*28+k)=arrtmp(k) Next flag=arrtmp(28) Next For i=1 To UBound(arrStim) TrialList.SetNested CStr(i), "list" & arrStim(i) Next TrialList.Reset 要求：有三种实验材料，A材料100个、BC材料各20个。BC不能呈现在首尾的前后两个刺激上。BC同种材料不能重复呈现，BC可以挨着呈现。对A材料不反应,反应位错；对B材料按“f”键进行反应，C材料按“j”键进行反应。实验过程为：注视点800毫秒→刺激界面(无限时间) →反馈界面。设计要求： 1）请用“化整为零+随机打乱判断”法进行设计。 对以上编码进行分析后改成一个新的编码

Dim arrStim(140), Typecount(3), arrtmp(28) As Integer
Dim i, k, flag As Integer
Dim bools As Boolean

'定义材料数量
Typecount(1) = 100 'A材料100个
Typecount(2) = 20 'B材料20个
Typecount(3) = 20 'C材料20个

'生成材料列表
For i = 1 To Typecount(1)
arrtmp(i) = 1 'A材料
Next

For i = Typecount(1) + 1 To Typecount(1) + Typecount(2) + Typecount(3)
If i = Typecount(1) + 1 Then '第一个BC材料
Do While True
arrtmp(i) = Int(Rnd() * 2) + 2 '随机生成2或3
If arrtmp(i) = 2 Then
flag = 2 '标记上一个材料是BC中的哪一个
Exit Do
ElseIf arrtmp(i) = 3 Then
flag = 3
Exit Do
End If
Loop
Else '第二个及以后的BC材料
Do While True
arrtmp(i) = Int(Rnd() * 2) + 2
If arrtmp(i) = 2 Then
If flag = 2 Then '如果上一个材料是B，重新生成
Continue Do
Else '否则标记为B
flag = 2
Exit Do
End If
ElseIf arrtmp(i) = 3 Then
If flag = 3 Then '如果上一个材料是C，重新生成
Continue Do
Else '否则标记为C
flag = 3
Exit Do
End If
End If
Loop
End If
Next

'随机打乱刺激顺序
RandomizeArray arrtmp, 1, UBound(arrtmp)

'将刺激列表按要求排列
For i = 1 To UBound(arrStim) Step 28
'首先判断BC不能呈现在首尾的前后两个刺激上
If i = 1 Then '第一个刺激
If arrtmp(i + 1) = 2 Or arrtmp(i + 1) = 3 Then
'如果第二个刺激是BC，将BC放到后面
For k = i + 2 To i + 27
If arrtmp(k) = 2 Or arrtmp(k) = 3 Then
arrtmp(i) = arrtmp(k)
arrtmp(k) = 1 '将BC换成A
Exit For
End If
Next
End If
ElseIf i = UBound(arrStim) - 27 Then '最后一个刺激
If arrtmp(i - 1) = 2 Or arrtmp(i - 1) = 3 Then
'如果倒数第二个刺激是BC，将BC放到前面
For k = i - 2 To i - 27 Step -1
If arrtmp(k) = 2 Or arrtmp(k) = 3 Then
arrtmp(i) = arrtmp(k)
arrtmp(k) = 1
Exit For
End If
Next
End If
Else '中间刺激
If arrtmp(i - 1) = 2 Or arrtmp(i - 1) = 3 Or arrtmp(i + 1) = 2 Or arrtmp(i + 1) = 3 Then
'如果前一个或后一个刺激是BC，将BC放到后面
For k = i + 2 To i + 27
If arrtmp(k) = 2 Or arrtmp(k) = 3 Then
arrtmp(i) = arrtmp(k)
arrtmp(k) = 1
Exit For
End If
Next
End If
End If

'然后判断BC同种材料不能重复呈现
For k = i To i + 27
If arrtmp(k) = 2 Then '当前是B
If k < i + 27 Then '不是最后一个
If arrtmp(k + 1) = 2 Then '后一个也是B
'将后面的第一个C换成A
For j = k + 1 To i + 27
If arrtmp(j) = 3 Then
arrtmp(j) = 1
Exit For
End If
Next
End If
End If
ElseIf arrtmp(k) = 3 Then '当前是C
If k > i Then '不是第一个
If arrtmp(k - 1) = 3 Then '前一个也是C
'将前面的第一个B换成A
For j = k - 1 To i Step -1
If arrtmp(j) = 2 Then
arrtmp(j) = 1
Exit For
End If
Next
End If
End If
End If
Next

'将刺激列表加入TrialList
For k = 1 To 28
arrStim(i + k - 1) = arrtmp(i + k - 1)
TrialList.SetNested CStr(i + k - 1), "list" & arrStim(i + k - 1)
Next
Next

TrialList.Reset

b=1 for i in range(1,20): if b>=10: break else: b+=3 continue b-=1 else: print(i,end=',') print(i)

As an AI language model, I can explain the code you provided.

The code is using a for loop to iterate over the range of numbers from 1 to 19. The variable b is initially set to 1.

Inside the loop, there is an if statement that checks if the value of b is greater than or equal to 10. If it is, then the loop is terminated using the break statement.

If the value of b is less than 10, then the else block is executed. In the else block, the value of b is set to 3 and the continue statement is used to skip the remaining statements in the loop and jump to the next iteration.

After the continue statement, the value of b is decremented by 1. The loop continues to iterate until it reaches the end of the range, or until the break statement is executed if b is greater than or equal to 10.

If the loop completes without encountering the break statement, then the else block is executed. In this case, the print statement prints the value of i followed by a comma, and the loop continues to the next iteration.

Finally, after the loop is completed, the print statement outside the loop prints the value of i again without a comma. However, since the loop was terminated early by the break statement, the value of i is not printed.

Therefore, the output of the code would be an empty string, as no value of i was printed.