vb for the topic of the loop control structure

When i=1, j=1, k=1 to 3, so a=a+1 is executed 3 times, a=+1+1+1=3

When i=2 and j=1, then k=1 to 3, a=a+1 is executed 3 more times, so a=3+3=6

j=2 then k=2 to 3, a=a+1 is executed 2 more times, so a=6+2=8

When i=3, j=1 then k=1 to 3, a=a+1 is executed 3 times, so a=8+3=11

j=2 then k=2 to 3, a=a+1 is executed 2 times, so a=11+2=13

j=3 then k=3 to 3, a=a+1 is executed 1 more time, so a=13+1=14

The last value of a is 14, which is to examine the number of executions of a=a+1 in this problem.

**, multiline=true for text1, scrollbar=2

private sub command1_click()for i = 1 to 3

for j = 1 to i

for k = j to 3

a = a + 1

vbcrlf & "i=" & i & ";j=" & j & ";k=" & k & ";a=a+1=" & a

next k

next j

next i

print a

end sub

The result is as follows:

i=1;j=1;k=1;a=a+1=1

i=1;j=1;k=2;a=a+1=2

i=1;j=1;k=3;a=a+1=3

i=2;j=1;k=1;a=a+1=4

i=2;j=1;k=2;a=a+1=5

i=2;j=1;k=3;a=a+1=6

i=2;j=2;k=2;a=a+1=7

i=2;j=2;k=3;a=a+1=8

i=3;j=1;k=1;a=a+1=9

i=3;j=1;k=2;a=a+1=10

i=3;j=1;k=3;a=a+1=11

i=3;j=2;k=2;a=a+1=12

i=3;j=2;k=3;a=a+1=13

i=3;j=3;k=3;a=a+1=14

Because the first two for loops will be x=3, it shouldn't need too much introduction, it's very good.

The key is the last for loop, 2 cycles from 1 to 2.

First cycle: x=3+5=8

Second cycle: x=8+5=12

So the final result is 13

I hope the landlord can understand and help the landlord.