This array shows music times: 1 - 1/2 - 1/4 - 1/8 and so on.
I think it should be written by recursive function, because I couldn't do it by while and for loop.
Now, I need all sorted figures that sum of them are equal 1. For example:
1/2 - 1/2
1/2 - 1/4 - 1/4
1/4 - 1/2 - 1/4
1/4 - 1/4 - 1/2
1/4 - 1/4 - 1/4 - 1/4
1/8 - 1/2 - 1/4 - 1/8
1/8 - 1/2 - 1/8 - 1/4
1/8 - 1/2 - 1/8 - 1/8 - 1/8
Of course, number of figures will be too many more if we add 1/16 and 1/32 and ... .
I can print them in a web page(response.write) or in a c console; it's not important. But the algorithm should be as simple as possible.
Thank you all.
HackerMan has a message in form of digits but he wants to decode the message so that if enemies gets hold of the message they will not be completely able to decode the message.
Since the message consists only of number, So decoding involves reversing the number. The first digit becomes last and vice versa. For example, if there is 1245 in the code, it becomes 5421 now.
Note : All the leading zeros are omitted. That means if the number ends with a zero, the zero is lost by reversing (e.g. 1200 gives 21).
HackerMan is further thinking of complicating the process and he needs your help. Your task is to add the numbers after reversing and output the result after reversing the sum.
The input consists of N cases. The first line of the input contains only positive integer N. Then follow the cases. Each case consists of exactly one line with two positive integers separated by space. These are the reversed numbers you are to add.
For each case, print exactly one line containing only one integer - the reversed sum of two reversed numbers. Omit any leading zeros in the output.
The value of N will be less than 10000.
The value of digits will be less than 5000.
When I run the command line app I can debug through the cpp dll into the C# dll but when I return from the cpp dll I get error on returning to the debug commanline app.
I get error:
An unhandled exception of type 'System.AccessViolationException' occurred in mscorlib.dll
Additional information: Attempted to read or write protected memory. This is often an indication that other memory is corrupt.
but i keep getting errors from
control.cpp, output.cpp, tessedit.cpp and tesseractemginewrapper.cpp
the error are all the same stating error C1010: unexpected end of file while looking for precompiled header. did you forget to add '#StdAfx.h' to your source?
as i am completely new i tried with/without the StdAfx.h and StdAfx.cpp which was there when i first created the project.
The message is telling you exactly what is wrong. Your project is set to use precompiled headers, but you forgot to include StdAfx.h in your source code. Each source file (.cpp file) should include the following line at the top:
When this event gets activated, does the event args structure get cloned for each object instance that subscribes to the event, or is there a single instance of the event object with each event handler getting a managed pointer to the same object?
What I'm really getting to is whether the large array gets cloned for each event handler. I'm guessing no.
Many thanks in advance for any pointers (no pun intended...)
The only other method to count 1 to 1000 is using recursion. According to C language, j has ‘1’as its value at the beginning. When 1 <= j < 1000, &main + (&exit - &main)*(j/1000) always evaluated to &main, which is the memory address of main. (&main)(j+1) is the next iteration we want to get, which would print ‘2’ on the screen, etc. The stop condition of this recursion is that When j hits 1000, &main + (&exit - &main)*(j/1000) evaluates to &exit, which will elegantly exit this process, and has the error code 1001 returned to the operating system.
is evaluated each time thus: &exit - &main is the offset from the start of main to the start of exit j/1000 will evaluate to zero for all vaules of j less than 1000
Multiplying those two values together gives zero if j is less than 1000.
Add that to &main (the address of main) and you get the same address, so main gets called with the parameter value j+1.
This continues until the value of j reaches 1000 at which time: j/1000 results in the value 1
That is multiplied by the offset of exit which returns that offset.
Add that value to main and the next call will go to exit rather than main.