C/C++ :: Prime Factorization Returns No Value
Feb 27, 2014
So, I have been working my way through lines of code, and practicing quite a bit. In fact, I have done programs far more complicated than this and still I cannot get it to work.
The idea of the code is to take a number input by the user through the terminal and check if it is a prime number(It does this) If the number is not prime it takes that number and checks for N amount of primes that is put in through the console. When a prime is found it writes that to the console until the N amount of primes have been found.
#include <iostream>
using namespace std;
int prime(unsigned long long x);
int main() {
long long x;
long long y;
long long z;
[Code] ....
At least that's my goal. Currently it will accept the number and check if it is prime. It will then ask for the N amount of primes but will not return any primes. I have check with numbers a I know have prime factors. I have added the comment //XXXXXXXX To show where the problem appears to occur.
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Dec 9, 2013
I am trying to factorize semi-prime numbers by using a brute force method and a square root method.The brute force method should start by multiplying all odd numbers between 1 and the semi-prime number. Meaning that 1 should multiply 3 then 5 then 7, etc. Then three should multiply 5 then 7 then 9 and so forth. The first number would never multiply a number equal to or less than itself.
bruteForce(int x){
f1 = 1;
f2 = f1 + 2;
while (f1 < x){
for (f2 = x; f2 <= x; f2-2){
[Code] ....
For my other algorithm, I want it to get the square root of the semi-prime and then start multiplying odd numbers around it. I made it so that the square root would have to be set to the next even number (in case it was a factor). If the product of the two odd numbers (a * b) is greater than the original semi-prime, I would multiply b by the next smaller number than a. Conversely, if the product of a and b is less than the original semi-prime, it will multiply a by the next odd number larger than b. This will repeat until the two correct factors of the original semi-prime are found.
Algorithm:
SquareRootFactorize(int x){
int N = x;
sqrtN = sqrt(N);
if (isOdd(sqrtN) == true)
sqrtN = sqrtN + 1;
f1 = sqrtN - 1;
f2 = sqrtN + 1;
[code].....
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Mar 12, 2014
the only thing is I want multiple all prime number I got. such as when I enter 12 the output should be "the prime factorization of 12 is 2*2*3" right now the program output is "2 2 3"
# include <stdio.h>
# include <math.h>
void primeFactors(int n)
{
[Code]....
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Feb 8, 2014
I wrote a function to do factorization (n!). while this function works perfectly:
Code:
int fact (double x){
register int z=1;
if (x==0)
return 1;
else
for(x;x;x--)
z*=x;
return z;
}
this doesn't:
Code:
int fact (double x){
if (x==0)
return 1;
else
for(x;x;x--)
x*=x;
return x;
}
what makes the second one different from the first one? can't see the difference..
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Dec 2, 2014
I have a program that gets prime numbers that doesn't work. It looks like it works for all prime numbers except odd non-prime ones. Here is the code:
Code:
#include <iostream>
using namespace std;
bool isPrime(int prime) {
while(true) {
int track = 0;
[Code] ...
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Aug 24, 2014
I'm a new coder for C++, and I've recently learned java before starting this programming language. I'm attempting to find all prime numbers up to the number a user enters (i.e. User enters 10, System shows "1,2,3,5,7"),
#include <iostream>
#include <cstdlib>
using namespace std;
int main(int argc, char** argv) {
int num;
cout << "Enter a positive number." << endl;
[Code] ....
I've been looking at my forloop, and I can't seem to pinpoint the problem with it.
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Apr 28, 2013
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Code:
#include <stdio.h>
#include <stdlib.h>
#include <mpi.h>
int isPrime(int num);
int main(int argc, char **argv){
}
[code]....
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class cylinderType {
public:
void getRadius(double rad);
void getHeight(double heigh);
double area();
double volume();
[code]....
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Code:
//this next line outputs correctly so I know my function is working
fprintf(stdout, "%s
", get_filename(selection));
char* temp;
[Code].....
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Write a C++ program consisting of main plus two other functions which will do the following:
Take an integer input from the keyboard.
Send the integer to a function which will output the integer to the screen.
Send the integer to a second function which will tell the user that the integer is an odd value.
Do not tell the user anything if the integer is an even value.
Repeat this process until the user enters something which is not an integer; use input validation to check for validity.
Any not valid input should terminate the program.
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bool setzen(int x, int y){
if(feld[x][y]!=3){
return mine=false;
}
[Code]....
Thats the Part where Everything happens but the Function "bool setzen" always returns false and i dont know why.
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DllExport double WINAPI GetSum(LPXLOPER12 arg)
{
if (arg->xltype == xltypeMissing) return -1;
return arg->xltype;
}
This code works as expected: if I miss argument, it gives me -1, and the type otherwise. But when I change code to this:
DllExport double WINAPI GetSum(LPXLOPER12 arg)
{
return (arg->xltype == xltypeMissing) ? -1 : arg->xltype;
}
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Code:
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else count = b.size();
[Code] .....
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Code:
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clock_gettime(CLOCK_MONOTONIC, &end);
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Code:
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char ret=0;
T *a_ = (a);
size_t n_ = (n);
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while(0)
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Dec 24, 2013
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float InputValidation(){
float num;
string strInput;
while (true){
getline(cin, strInput);
[Code] ....
You also need to include <string> and <sstream>.
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Code:
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[Code] ....
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if (lo > hi) {
return 0;
}
if (lo == hi)
[Code]...
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Jan 26, 2014
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[Code] .....
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the same is true for this example:
class Program {
static void printvalues(params int[] passedin) {
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Console.WriteLine(printthis);
[Code] ....
This code won't compile without the static keyword in the printvalues() method signature. why?
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