Question

执行下面的伪代码，输出的结果是

I←1
For n From 1 To 11 Step 2
I←2I+1
If I>20 Then
I←I-20
End if
End For
Print I

Answer 1

#include <stdio.h>

int main()
{
int i=1;
int n;
for(n=1;n<=11;n+=2)
{
i=2*i+1;
if(i>20)
{
i-=20;
}
}
printf("%d",i);
return 0;
}

答案是7

Answer 2

结果是7
int main()
{int i=1;
for(int n=1;n<12;n+=2)
{
i=2*i+1;
if(i>20)
i=i-20;
}

cout<<i;
return 0;
}
一楼的真快啊~

Answer 3

i = 1
For n = 1 To 11 Step 2
i = 2 * i + 1
If i > 20 Then
i = i - 20
End If
Next
Print i
结果为7