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  • how to write the code for this program specially in mathematica? [closed]

    - by asd
    I implemented a solution to the problem below in Mathematica, but it takes a very long time (hours) to compute f of kis or the set B for large numbers. Somebody suggested that implementing this in C++ resulted in a solution in less than 10 minutes. Would C++ be a good language to learn to solve these problems, or can my Mathematica code be improved to fix the performance issues? I don't know anything about C or C++ and it should be difficult to start to learn this languages. I prefer to improve or write new code in mathematica. Problem Description Let $f$ be an arithmetic function and A={k1,k2,...,kn} are integers in increasing order. Now I want to start with k1 and compare f(ki) with f(k1). If f(ki)f(k1), put ki as k1. Now start with ki, and compare f(kj) with f(ki), for ji. If f(kj)f(ki), put kj as ki, and repeat this procedure. At the end we will have a sub sequence B={L1,...,Lm} of A by this property: f(L(i+1))f(L(i)), for any 1<=i<=m-1 For example, let f is the divisor function of integers. Here I put some part of my code and this is just a sample and the question in my program could be more larger than these: «««««««««««««««««««««««««««««««««««« f[n_] := DivisorSigma[0, n]; g[n_] := Product[Prime[i], {i, 1, PrimePi[n]}]; k1 = g[67757] g[353] g[59] g[19] g[11] g[7] g[5]^2 6^3 2^7; k2 = g[67757] g[353] g[59] g[19] g[11] g[7] g[5] 6^5 2^7; k3 = g[67757] g[359] g[53] g[19] g[11] g[7] g[5] 6^4 2^7; k4 = g[67759] g[349] g[53] g[19] g[11] g[7] g[5] 6^5 2^6; k5 = g[67757] g[359] g[53] g[19] g[11] g[7] g[5] 6^4 2^8; k6 = g[67759] g[349] g[53] g[19] g[11] g[7] g[5]^2 6^3 2^7; k7 = g[67757] g[359] g[53] g[19] g[11] g[7] g[5] 6^5 2^6; k8 = g[67757] g[359] g[53] g[19] g[11] g[7] g[5] 6^4 2^9; k9 = g[67757] g[359] g[53] g[19] g[11] g[7] g[5]^2 6^3 2^7; k10 = g[67757] g[359] g[53] g[19] g[11] g[7] g[5] 6^5 2^7; k11 = g[67759] g[349] g[53] g[19] g[11] g[7] g[5]^2 6^4 2^6; k12 = g[67757] g[359] g[53] g[19] g[11] g[7] g[5]^2 6^3 2^8; k13 = g[67757] g[359] g[53] g[19] g[11] g[7] g[5]^2 6^4 2^6; k14 = g[67757] g[359] g[53] g[19] g[11] g[7] g[5]^2 6^3 2^9; k15 = g[67757] g[359] g[53] g[19] g[11] g[7] g[5]^2 6^4 2^7; k16 = g[67757] g[359] g[53] g[23] g[11] g[7] g[5] 6^4 2^8; k17 = g[67757] g[359] g[59] g[19] g[11] g[7] g[5] 6^4 2^7; k18 = g[67757] g[359] g[53] g[23] g[11] g[7] g[5] 6^4 2^9; k19 = g[67759] g[353] g[53] g[19] g[11] g[7] g[5] 6^4 2^6; k20 = g[67763] g[347] g[53] g[19] g[11] g[7] g[5] 6^4 2^7; k = Table[k1, k2, k3, k4, k5, k6, k7, k8, k9, k10, k11, k12, k13, k14, k15, k16, k17, k18, k19, k20]; i = 1; count = 0; For[j = i, j <= 20, j++, If[f[k[[j]]] - f[k[[i]]] > 0, i = j; Print["k",i]; count = count + 1]]; Print["count= ", count] ««««««««««««««««««««««««««««««««««««

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  • unformatted input to a std::string instead of c-string from binary file.

    - by posop
    ok i have this program working using c-strings. I am wondering if it is possible to read in blocks of unformatted text to a std::string? I toyed arround with if >> but this reads in line by line. I've been breaking my code and banging my head against the wall trying to use std::string, so I thought it was time to enlist the experts. Here's a working program you need to supply a file "a.txt" with some content to make it run. i tried to fool around with: in.read (const_cast<char *>(memblock.c_str()), read_size); but it was acting odd. I had to do std::cout << memblock.c_str() to get it to print. and memblock.clear() did not clear out the string. anyway, if you can think of a way to use STL I would greatly appreciate it. Here's my program using c-strings // What this program does now: copies a file to a new location byte by byte // What this program is going to do: get small blocks of a file and encrypt them #include <fstream> #include <iostream> #include <string> int main (int argc, char * argv[]) { int read_size = 16; int infile_size; std::ifstream in; std::ofstream out; char * memblock; int completed = 0; memblock = new char [read_size]; in.open ("a.txt", std::ios::in | std::ios::binary | std::ios::ate); if (in.is_open()) infile_size = in.tellg(); out.open("b.txt", std::ios::out | std::ios::trunc | std::ios::binary); in.seekg (0, std::ios::beg);// get to beginning of file while(!in.eof()) { completed = completed + read_size; if(completed < infile_size) { in.read (memblock, read_size); out.write (memblock, read_size); } // end if else // last run { delete[] memblock; memblock = new char [infile_size % read_size]; in.read (memblock, infile_size % read_size + 1); out.write (memblock, infile_size % read_size ); } // end else } // end while } // main if you see anything that would make this code better please feel free to let me know.

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  • How do you compile a resource into the binary?

    - by Wonderflonium
    How do you compile a resource into the binary in XCode? That way is doesn't show up inside the application bundle as a file and it's not subject to manipulation by a user (whether good-intentioned or bad). Is this even possible? I'm particularly interested about this in terms of iPhone apps. Any help is appreciated!

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  • How to convert binary, OCT, HEX to calculate in Java?

    - by user316751
    Write an application that inputs one number consisting of FIVE digits from the user, separates the number into its individual digits and prints the digits separated from one another by three spaces each. For example, if the user types in the number 12345, the program should print 1 2 3 4 5 The following screen dump of result is for your reference. Input a digit: 12345 Digits in 12345 = 1 2 3 4 5 How to convert binary, OCT, HEX to calculate the question?

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  • How do I make a Mach-O binary from an object file?

    - by Mike
    I have an object file, which I'd like to make into a Mach-O binary(I'm on Mac OS 10.6). Running ld source.o produces the following output: Undefined symbols: "_printf", referenced from: _main in source.o ld: symbol(s) not found for inferred architecture x86_64 How can I reference libSystem(or whatever library includes printf) using ld?

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  • How can I match a match a null byte (0x00) in the Visual Studio binary editor with a find using a re

    - by Paul K
    Open a file in the Visual Studio binary editor that contains a null byte (0x00), then use the Quick Find feature (Ctrl +F) to find null bytes. I would have thought I could use a regular expression such as \x00 to match null bytes but it doesn't work. Searching for any other hex value using this method works fine. Is this a VS bug, 'feature', or am I just missing something? Is there a work around?

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  • How do I embed a binary within a Delphi executable and extract at runtime?

    - by Mick
    I'd like to embed an executable inside of my Delphi binary and extract it at runtime. The purpose of this is to ensure that a helper utility is always available on the system, without having to distribute multiple files. With Delphi 2007 I have used JvDataEmbedded, but I am building a console application and I want to know if anyone knows of another way to do it without having to add a hidden form for JvDataEmbedded. I am using Delphi 2010.

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  • using LARGE_INTEGER gives me back error error C2679: '=' binary no operator found which takes a right-hand operand

    - by rekotc
    i have the following code: QueryPerformanceCounter(&timeStart); winMain::render(); //do stuff QueryPerformanceCounter(&timeEnd); numCounts = ( timeEnd.QuadPart - timeStart.QuadPart); All the 3 variables are declared as LARGE_INTEGER, the code should work since im following a book example, but i get: error C2679: '=' binary no operator found which takes a right-hand operand of type LONGLONG it might be '_LARGE_INTEGER &_LARGE_INTEGER::operator =(const _LARGE_INTEGER &)' 1 durante la ricerca di corrispondenza con l'elenco di argomenti '(LARGE_INTEGER, LONGLONG)'

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  • C# How can i remove newline characters from binary?

    - by Tom
    Basically i have binary data, i dont mind if it's unreadable but im writing it to a file which is parsed and so it's importance newline characters are taken out. I thought i had done the right thing when i converted to string.... byte[] b = (byte[])SubKey.GetValue(v[i]); s = System.Text.ASCIIEncoding.ASCII.GetString(b); and then removed the newlines String t = s.replace("\n","") but its not working ?

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  • TrueCrypt with ext2/3 partition write access under Mac OS X Snow Leopard ?

    - by ssc
    I'm using a TrueCrypt volume with an ext3 partition under Snow Leopard with MacFUSE. I can mount ordinary (unencrypted) ext3 partitions read/write from the shell by adding command line arguments as shown in "Mounting ext3 in Snow Leopard…". However, TrueCrypt mounts the partition read-only and I don't see any way to 'sneak in' the required additional arguments. How do I mount it read/write? I was hoping for a similar solution as for mounting NTFS, but diskutil info /Volumes/my_volume/ does not return a UUID; it does tell me Read-Only Media: No Read-Only Volume: Yes though...

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