Write a function in VB

Parameter description: rndnumcount, the number of 6 digits; mysearchnum, the number to look for.

private function searchnumcount(byval rndnumcount as integer, byval mysearchnum as integer) as integer

dim i as integer '6-digit number.

dim j as long 'A random 6-digit value.

dim a as integer 'The 6-digit number of the current query.

dim start as integer 'The starting location of the query.

dim temp as integer 'Temporary location.

dim length as integer'Look at the length of the number (1).

redim marray(rndnumcount - 1) 'Define an array of parameter lengths.

randomize 'Random number seeds.

for i = 0 to rndnumcount - 1 'Assign a random 6-digit number to an array.

j = clng(999999 * rnd + 100000)

marray(i) = j

nextfor a=0 to rndnumcount-1 'The number of occurrences of the number to be queried in the array.

start=1

length=len(mysearchnum)

do temp=instr(start,marray(a),mysearchnum)

if temp<>0 then

searchnumcount = searchnumcount + 1

start=temp+length

else exit do

end if

loop next

msgbox "The number you're looking for" +cstr(mysearchnum)+", appears"+cstr(rndnumcount)+"The number of 6-digit numbers is:"+cstr(searchnumcount)+"Times! "

end function

dim num as long

num=0for i=1 to n

for j=0 to 5

if array(j) = the number you are looking for then num=num+1next j

next i

The value of num is the number of occurrences you are looking for.

You first find the number of times a value appears in each array, and then add these values together, and that's it, to provide you with a little idea, which you should be able to do by looking at the VB tutorial.

1，2，3，4...9 iterate over n arrays in turn.

This algorithm is a bit slow, and you can look up some about finding algorithms!

Write more**. Do a lot of online research. Ask others how to achieve it, and it will definitely be no problem, VB is very simple!