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Thread Rules
1. This is not a "do my homework for me" thread. If you have specific questions, ask, but don't post an assignment or homework problem and expect an exact solution.
2. No recruiting for your cockamamie projects (you won't replace facebook with 3 dudes you found on the internet and $20)
3. If you can't articulate why a language is bad, don't start slinging shit about it. Just remember that nothing is worse than making CSS IE6 compatible.
4. Use [code] tags to format code blocks. |
On October 28 2012 03:20 Thorakh wrote:+ Show Spoiler +private static boolean containsDuplicates(int[] array)
{
for (int i=0; i<array.length; i++)
{
if (array[i] != 0)
{
for (int j=i+1; j<array.length; j++)
{
if (array[i] == array[j]
{
return true;
}
}
}
}
return false;
} What on earth would cause this code to hang? I'm trying to check an array for duplicate integers, ignoring zeroes. Sometimes it hangs, sometimes it doesn't O_O
It shouldn't hang at any point but it will throw an exception when i == array.length - 1 if i read this correctly because then j == array.length => out of bounds.
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On October 27 2012 18:04 supereddie wrote:
You should be able to figure out what is the bottleneck in your code. If there is no clear bottleneck and the CPU is not running at 100% capacity, it might be possible to use more than one thread to do the processing (maybe one thread per line, max of 10 threads). But perhaps this is not allowed when using Java stored procedures.
I have no way of monitoring CPU load.
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On October 28 2012 03:23 Morfildur wrote:Show nested quote +On October 28 2012 03:20 Thorakh wrote:+ Show Spoiler +private static boolean containsDuplicates(int[] array)
{
for (int i=0; i<array.length; i++)
{
if (array[i] != 0)
{
for (int j=i+1; j<array.length; j++)
{
if (array[i] == array[j]
{
return true;
}
}
}
}
return false;
} What on earth would cause this code to hang? I'm trying to check an array for duplicate integers, ignoring zeroes. Sometimes it hangs, sometimes it doesn't O_O It shouldn't hang at any point but it will throw an exception when i == array.length - 1 if i read this correctly because then j == array.length => out of bounds. It isn't doing that though, and I haven't caught exceptions anywhere in my code.
But now that you mention it, that certainly is weird, especially the fact that no exceptions are being thrown.
edit: no wait, it never throws an exception because the code that would cause array[i] == array[j] to be an array overflow never executes when j=i+1 is larger than the array length.
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Add debugging statements and exception handlers and see where it hangs. From that you can figure out your next step.
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On October 28 2012 02:06 ArdentZeal wrote:
Hello there,
i could need some help in c++.
I have several classes (State1, State2, State3, ...) derived from the base class State.
They all have the same method, which is used sligthly differently depending on the State i am in.
So I wrote the method Automat::setCurrentState(State* _nextState); to set the States accordingly.
But if i try to call this method with one of the derived classes, i get a compiler error. Do i really have to overwrite Automat::setCurrentState(State* _nextState); with each of the 20 States or is there a simple solution i am missing?
Some additional info might be useful. What is the compiler error?
Just a wild guess: are you using private or protected inheritance?
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On October 27 2012 16:18 Craton wrote:Oracle is housed on a secure server and we simply call whatever output procedure to do its thing (it dumps it to the filesystem of said secure server). Data CANNOT leave secure servers. Even moving from one server to the next requires a secure, semi-automated queuing system to move any files.
Because of the complexity of the data and what we do with it, as well as how much is already designed around Oracle, it makes no sense to try and redo everything in an environment outside of Oracle.
If you want to answer my question, then do so. Stop fighting me on existing business processes.
All I wanted to know is if I could match or better the output speed of what I have now versus a Java procedure that accomplished the same task more cleanly. Trying to point at everything EXCEPT what I asked is extremely aggravating. Ok then, that's really unfortunate. I'm sorry if you felt I was fighting on this point, but from an outside perspective it is extremely aggravating to have to optimize the I/O bound last tiny bit of a problem, instead of a giant juicy easily-optimizable pile in the middle.
Now that I know you're limited to a secure room, oh well :\ At least you don't have to do your programming in an isolated, soundproofed, 200lbs-door locked room with no internet. Could be worse than simply having the unfortunate constraint of orcale and a single machine.
On October 27 2012 16:18 Craton wrote:I wouldn't even rate this as Alpha currently, so a lot of things are run step by step to bring things in and then modify it. The actual processing time for reading things in and then modifying them is pretty low (much lower than the time to output). So like last post, how much time does it take to just read the data you need out of orcale as fast as possible? (Don't even save, just read and drop - maybe do something uninteresting to make sure the compiler doesn't optimize out). Once you know how fast your I/O is, you know the lower bound on how fast you can get. Figure out what the timing on your existing operation is, is it an order of magnitude worse? If so, we have something to work on, if not, I'd just throw up my hands and move on to the next problem.
If it is an order of magnitude worse, can you step through with a profiler? There's java visual vm, and a number of open source tools. I'm not sure about attaching it to an oracle db stored procedure, but you should be able to replicate the behavior with a not-in-orcale plain old java program.
Or there is supposedly a stored procedure profiler, but I'm not sure you're going to be able to get cpu / stack trace timing out of it, which is what we're really after - that is, unless the first thing you see after loading up the profiler is massive thrashing.
http://docs.oracle.com/cd/B10501_01/appdev.920/a96624/12_tune.htm#45936
On October 28 2012 03:24 Craton wrote:Show nested quote +On October 27 2012 18:04 supereddie wrote:
You should be able to figure out what is the bottleneck in your code. If there is no clear bottleneck and the CPU is not running at 100% capacity, it might be possible to use more than one thread to do the processing (maybe one thread per line, max of 10 threads). But perhaps this is not allowed when using Java stored procedures. I have no way of monitoring CPU load.
shit just saw this, even with a profiler? Even with a profiler on a normal java program outside of oracle?
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Java
Alright, my problem is bigger than I thought.
The following recursive method attempts to solve a sudoku (note that it works if there are no duplicates present in the supplied grid):
+ Show Spoiler +public static boolean hasSolution(int[] grid)
{
boolean solved = false;
int candidateCell;
int trialValue;
if (isFull(grid))
{
solved = true;
}
else
{
candidateCell = getEmptyCell(grid);
trialValue = 1;
while (!solved && trialValue <= 9)
{
if (isLegal(grid, candidateCell, trialValue))
{
setCell(grid, candidateCell, trialValue);
if (hasSolution(grid))
{
solved = true;
}
else
{
clearCell(grid, candidateCell);
}
}
trialValue++;
}
}
return solved;
}
isFull(), getEmptyCell() - get's the next empty cell (a cell containing a 0), setCell() and clearCell() - resets the cell back to 0 - speak for themselves. The problem appears to be in isLegal():
+ Show Spoiler +private static boolean isLegal(int[] grid, int candidateCell, int trialValue)
{
//Check the row of candidateCell for duplicate numbers. If a duplicate is found, the solution is illegal.
//Calculate the row number of the candidate cell.
int rowNumber = candidateCell / 9;
//Store the values of the row corresponding to the calculated row number in a new array.
int[] row = new int[9];
for (int i=0; i<row.length; i++)
{
row[i] = grid[rowNumber * 9 + i];
}
//Insert the trial value at the correct position.
row[candidateCell % 9] = trialValue;
//Check the row for duplicates, ignoring duplicate 0's.
if (containsDuplicates(row))
{
return false;
}
return true;
}
with containsDuplicates() being the code I posted earlier. Note that isLegal() only checks for rows at the moment. I have code for columns and blocks as well but I first need to solve this. I tried some sample grids to check if the code works by doing:
+ Show Spoiler +int[] grid = new int[] {0, 0, 0, 0, 0, 0, 0, 0, 0,
1, 0, 1, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0};
if (SudokuSolver.hasSolution(grid))
{
SudokuSolver.showGrid(grid);
}
else
{
System.out.println("No solution!");
}
Now, this sample grid works perfectly. It has no solution and prints "No solution!". Sample grids with multiple 1's on one of the first two rows work (they tell me there is no solution). However, sample grids with multiple 1's on the 3th, 4th, 5th, 6th, 7th, 8th or 9th row, like:
+ Show Spoiler +int[] grid = new int[] {0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 1, 0, 1, 0, 0, 1, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0};
will keep running forever for an inexplicable reason. What the hell is happening here? I'm completely stumped.
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On October 28 2012 03:23 Morfildur wrote:Show nested quote +On October 28 2012 03:20 Thorakh wrote:+ Show Spoiler +private static boolean containsDuplicates(int[] array)
{
for (int i=0; i<array.length; i++)
{
if (array[i] != 0)
{
for (int j=i+1; j<array.length; j++)
{
if (array[i] == array[j]
{
return true;
}
}
}
}
return false;
} What on earth would cause this code to hang? I'm trying to check an array for duplicate integers, ignoring zeroes. Sometimes it hangs, sometimes it doesn't O_O It shouldn't hang at any point but it will throw an exception when i == array.length - 1 if i read this correctly because then j == array.length => out of bounds.
It seems to be right, the problem must be elsewhere in your code.
package frigo;
public class DuplicateFinder {
public static boolean containsDuplicates (int[] array) {
for( int i = 0; i < array.length; i++ ){
if( array[i] != 0 ){
for( int j = i + 1; j < array.length; j++ ){
if( array[i] == array[j] ){
return true;
}
}
}
}
return false;
}
}
package frigo;
import static frigo.DuplicateFinder.containsDuplicates;
import static org.hamcrest.Matchers.is;
import static org.junit.Assert.assertThat;
import org.junit.Test;
public class DuplicateFinderTest {
@Test
public void testEmptyArrayDoesNotContainDuplicates () {
int[] array = {};
assertThat(containsDuplicates(array), is(false));
}
@Test
public void testMultipleElementArrayContainsDuplicates () {
int[] array = {0, 1, 1, 3, 4};
assertThat(containsDuplicates(array), is(true));
}
@Test
public void testMultipleELementArrayDoesNotContainDuplicates () {
int[] array = {0, 1, 2, 3, 4};
assertThat(containsDuplicates(array), is(false));
}
@Test
public void testSingleElementArrayDoesNotContainDuplicates () {
int[] array = {1};
assertThat(containsDuplicates(array), is(false));
}
@Test
public void testZerosAreNotCountedAsDuplicates () {
int[] array = {0, 0};
assertThat(containsDuplicates(array), is(false));
}
}
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On October 28 2012 03:50 Thorakh wrote:JavaAlright, my problem is bigger than I thought. The following recursive method attempts to solve a sudoku (note that it works if there are no duplicates present in the supplied grid): + Show Spoiler +public static boolean hasSolution(int[] grid)
{
boolean solved = false;
int candidateCell;
int trialValue;
if (isFull(grid))
{
solved = true;
}
else
{
candidateCell = getEmptyCell(grid);
trialValue = 1;
while (!solved && trialValue <= 9)
{
if (isLegal(grid, candidateCell, trialValue))
{
setCell(grid, candidateCell, trialValue);
if (hasSolution(grid))
{
solved = true;
}
else
{
clearCell(grid, candidateCell);
}
}
trialValue++;
}
}
return solved;
} isFull(), getEmptyCell() - get's the next empty cell (a cell containing a 0), setCell() and clearCell() - resets the cell back to 0 - speak for themselves. The problem appears to be in isLegal(): + Show Spoiler +private static boolean isLegal(int[] grid, int candidateCell, int trialValue)
{
//Check the row of candidateCell for duplicate numbers. If a duplicate is found, the solution is illegal.
//Calculate the row number of the candidate cell.
int rowNumber = candidateCell / 9;
//Store the values of the row corresponding to the calculated row number in a new array.
int[] row = new int[9];
for (int i=0; i<row.length; i++)
{
row[i] = grid[rowNumber * 9 + i];
}
//Insert the trial value at the correct position.
row[candidateCell % 9] = trialValue;
//Check the row for duplicates, ignoring duplicate 0's.
if (containsDuplicates(row))
{
return false;
}
return true;
} with containsDuplicates() being the code I posted earlier. Note that isLegal() only checks for rows at the moment. I have code for columns and blocks as well but I first need to solve this. I tried some sample grids to check if the code works by doing: + Show Spoiler +int[] grid = new int[] {0, 0, 0, 0, 0, 0, 0, 0, 0,
1, 0, 1, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0};
if (SudokuSolver.hasSolution(grid))
{
SudokuSolver.showGrid(grid);
}
else
{
System.out.println("No solution!");
}Now, this sample grid works perfectly. It has no solution and prints "No solution!". Sample grids with multiple 1's on one of the first two rows work (they tell me there is no solution). However, sample grids with multiple 1's on the 3th, 4th, 5th, 6th, 7th, 8th or 9th row, like: + Show Spoiler +int[] grid = new int[] {0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 1, 0, 1, 0, 0, 1, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0};will keep running forever for an inexplicable reason. What the hell is happening here? I'm completely stumped.
Seems like you need a 2d array bro.
Your position tracking seems pretty hacky to me and prone to bugs.
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Hmm, I just rewrote my program and it seems to work perfectly now, instead of doing all kinds of position calculations I just check the entire grid for legality of a move every time:
+ Show Spoiler +public class Sudoku
{
public static boolean hasSolution(int[] grid)
{
boolean solved = false;
int candidateCell;
int trialValue;
if (!isGridLegal(grid))
{
return false;
}
if (isFull(grid))
{
solved = true;
}
else
{
candidateCell = getEmptyCell(grid);
trialValue = 1;
while (!solved && trialValue <= 9)
{
setCell(grid, candidateCell, trialValue);
if (hasSolution(grid))
{
solved = true;
}
else
{
clearCell(grid, candidateCell);
}
trialValue++;
}
}
return solved;
}
public static void showGrid(int[] grid)
{
for (int i=0; i<grid.length; i++)
{
System.out.print(grid[i]+" ");
if ((i+1) % 3 == 0)
{
System.out.print(" ");
}
if ((i+1) % 27 == 0)
{
System.out.println();
}
if ((i+1) % 9 == 0)
{
System.out.println();
}
}
}
private static boolean isGridLegal(int[] grid)
{
return areRowsLegal(grid) && areColumnsLegal(grid) && areBlocksLegal(grid);
}
private static boolean areRowsLegal(int[] grid)
{
int row[] = new int[9];
for (int i=0; i<grid.length; i+=9)
{
for (int j=0; j<9; j++)
{
row[j] = grid[i + j];
}
if (containsDuplicates(row))
{
return false;
}
}
return true;
}
private static boolean areColumnsLegal(int[] grid)
{
int[] column = new int[9];
for (int i=0; i<9; i++)
{
for (int j=0; j<grid.length; j+=9)
{
column[j / 9] = grid[i + j];
}
if (containsDuplicates(column))
{
return false;
}
}
return true;
}
private static boolean areBlocksLegal(int[] grid)
{
int[][] blocks = new int[9][9];
blocks[0] = new int[] { 0, 1, 2, 9, 10, 11, 18, 19, 20};
blocks[1] = new int[] { 3, 4, 5, 12, 13, 14, 21, 22, 23};
blocks[2] = new int[] { 6, 7, 8, 15, 16, 17, 24, 25, 26};
blocks[3] = new int[] {27, 28, 29, 36, 37, 38, 45, 46, 47};
blocks[4] = new int[] {30, 31, 32, 39, 40, 41, 48, 49, 50};
blocks[5] = new int[] {33, 34, 35, 42, 43, 44, 51, 52, 53};
blocks[6] = new int[] {54, 55, 56, 63, 64, 65, 72, 73, 74};
blocks[7] = new int[] {57, 58, 59, 66, 67, 68, 75, 76, 77};
blocks[8] = new int[] {60, 61, 62, 69, 70, 71, 78, 79, 80};
for (int i=0; i<blocks.length; i++)
{
int block[] = new int[9];
for (int j=0; j<blocks[i].length; j++)
{
block[j] = grid[blocks[i][j]];
}
if (containsDuplicates(block))
{
return false;
}
}
return true;
}
private static boolean containsDuplicates(int[] array)
{
for (int i=0; i<array.length; i++)
{
if (array[i] != 0)
{
for (int j=i+1; j<array.length; j++)
{
if (array[i] == array[j])
{
return true;
}
}
}
}
return false;
}
private static void clearCell(int[] grid, int cell)
{
grid[cell] = 0;
}
private static void setCell(int[] grid, int cell, int value)
{
grid[cell] = value;
}
private static int getEmptyCell(int[] grid)
{
for (int i=0; i<grid.length; i++)
{
if (grid[i] == 0)
{
return i;
}
}
return -1;
}
private static boolean isFull(int[] grid)
{
for (int i=0; i<grid.length; i++)
{
if (grid[i] == 0)
{
return false;
}
}
return true;
}
}
edit: changed hasSolution() a bit because if the very last cell in the grid would be open, it would just insert a 1 and isFull() would return true and report the sudoku solved without checking if the 1 in the last cell actually solved it.
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1019 Posts
Can you guys help me? I need to write a simple program where I have to store a user-inputted number of primes into a vector. Generating the number of primes is easy but I don't know how to make it so that it only stores X number of vectors :/
Here is my code so far. It outputs nothing when I enter an input for NPrimes, like NPrimes = 10.
+ Show Spoiler +
cout << "Input the number of primes to find: ";
int NPrimes;
cin >> NPrimes;
vector<int> primes;
int k = 0;
for (int i = 2; i < 1000000; i++) //checking for all primes between 2 and 100000
{
bool isprime = true;
for (int j = 2; j <= sqrt(i); j++)
{
if (i%j == 0)
{
isprime = false;
}
}
if (isprime == true) // If i is a prime, it goes into the do-while loop and plugs in each i into the vector
{ // "primes" for "NPrimes" number of times.
do
{
primes.push_back(i);
k++;
}while (k < NPrimes);
}
}
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On October 29 2012 01:50 white_horse wrote:
do
{
primes.push_back(i);
k++;
}while (k < NPrimes);
That's just plain silly.
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On October 29 2012 01:50 white_horse wrote:Can you guys help me? I need to write a simple program where I have to store a user-inputted number of primes into a vector. Generating the number of primes is easy but I don't know how to make it so that it only stores X number of vectors :/ Here is my code so far. It outputs nothing when I enter an input for NPrimes, like NPrimes = 10. + Show Spoiler +
cout << "Input the number of primes to find: ";
int NPrimes;
cin >> NPrimes;
vector<int> primes;
int k = 0;
for (int i = 2; i < 1000000; i++) //checking for all primes between 2 and 100000
{
bool isprime = true;
for (int j = 2; j <= sqrt(i); j++)
{
if (i%j == 0)
{
isprime = false;
}
}
if (isprime == true) // If i is a prime, it goes into the do-while loop and plugs in each i into the vector
{ // "primes" for "NPrimes" number of times.
do
{
primes.push_back(i);
k++;
}while (k < NPrimes);
}
}
Don't have time to figure out exactly why your code doesn't work, but it's quite bloated and I think I see a logic error in your do while. Try something simpler (I did not test this code), note that j * j < i is the same as i < sqrt( j ), except it's much faster due to avoiding the slow sqrt( ) function:
cout << "Input the max range to search: ";
int range;
cin >> range;
vector<int> primes;
for(int i = 0; i < range; ++i)
{
for(int j = 2; j * j < i; ++j)
{
if(i % j == 0)
{
primes.push_back( j );
}
}
}
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On October 29 2012 02:59 Frigo wrote:
do
{
primes.push_back(i);
k++;
}while (k < NPrimes);
That's just plain silly.
Yeah this is broken.
Don't have time to figure out exactly why your code doesn't work, but it's quite bloated and I think I see a logic error in your do while. Try something simpler (I did not test this code), note that j * j < i is the same as i < sqrt( j ), except it's much faster due to avoiding the slow sqrt( ) function: cout << "Input the max range to search: ";
int range;
cin >> range;
vector<int> primes;
for(int i = 0; i < range; ++i)
{
for(int j = 2; j * j < i; ++j)
{
if(i % j == 0)
{
primes.push_back( j );
}
}
}
You sure about (j * j < i) being the same?
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On October 29 2012 03:17 Fyodor wrote:
You sure about (j * j < i) being the same?
Well, take a looky:
j < sqrt( i )
square each side
j^2 < i
therefor
j * j < i
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On October 29 2012 03:28 CecilSunkure wrote:Show nested quote +On October 29 2012 03:17 Fyodor wrote:
You sure about (j * j < i) being the same? Well, take a looky: j < sqrt( i )
square each side
j^2 < i
therefor
j * j < i
Fascinating. You're good.
BTW, whitehorse, if you want to stop a loop after X passes or at a certain point... you need a break statement.
Like in your original code, you use:
if (k >= NPrimes)
{
break;
}
if you want the program to stop once it found the right amount of primes. Your while loop won't loop only once every pass, it'll loop like 20 times and push the same value over and over.
|
1019 Posts
we're not allowed to use breaks or continues  I still don't understand how to store, say just 10 prime numbers in the vector.
|
On October 29 2012 07:08 white_horse wrote:we're not allowed to use breaks or continues I still don't understand how to store, say just 10 prime numbers in the vector.
Well it will find all the primes unless you break out of the division test.
You could make the loop index break the loop condition on purpose maybe? It wouldn't be an explicit break statement but it would achieve the same thing.
cout << "Input the max range to search: ";
int range;
cin >> range;
vector<int> primes;
for(int i = 0; i < range; ++i)
{
for(int j = 2; j * j < i; ++j)
{
if(i % j == 0)
{
primes.push_back( j );
}
}
if (primes.size() >= 10) // breaks the loop after it finds 10 prime numbers.
{
i = range;
}
}
(only works if size() returns an int lol, not sure if it does)
|
On October 29 2012 07:22 Fyodor wrote:Show nested quote +On October 29 2012 07:08 white_horse wrote:we're not allowed to use breaks or continues I still don't understand how to store, say just 10 prime numbers in the vector. Well it will find all the primes unless you break out of the division test. You could make the loop index break the loop condition on purpose maybe? It wouldn't be an explicit break statement but it would achieve the same thing.
cout << "Input the max range to search: ";
int range;
cin >> range;
vector<int> primes;
for(int i = 0; i < range; ++i)
{
for(int j = 2; j * j < i; ++j)
{
if(i % j == 0)
{
primes.push_back( j );
}
}
if (primes.size() >= 10) // breaks the loop after it finds 10 prime numbers.
{
i = range;
}
}
(only works if size() returns an int lol, not sure if it does)
This won't break out of the second loop properly and you'll end up a bunch of primes. white_horse you just need to make your loop end when the vector's size reaches something. Where can you do this? In the comparison of the for loop using &&.
cout << "Input the max range to search: ";
int range;
cin >> range;
vector<int> primes;
const unsigned SIZE = 10;
for(int i = 0; i < range && primes.size( ) != SIZE; ++i)
{
for(int j = 2; j * j < i && primes.size( ) != SIZE; ++j)
{
if(i % j == 0)
{
primes.push_back( j );
}
}
}
There are lots of ways to detect when you have reached ten. You can use primes.size( ), you can count with an int each time you use push back.
There are also lots of ways to end your nested for loop; you can use conditions like I did, you could place your for loops into a function and use return to break both loops (probably the best way), you can use the goto trick to break from nested for loops, you can place the push_back call in an if statement and only push back if the size is under the SIZE constant (doesn't break the loop but stops adding), etc.
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EPT
