Security setting problem!! need help

i m a designer ,once i done i need to email  the pdf file to cilent .But i wanted to make the file with security !! i just want my cilent just have a view but cant print and save on that file beacuse i scare the cilent use my design as a copyright .
so anyone here can teach me how to set the security mode to cant pirnt and save?can teach me step by step? i using adobe reader X and normally i using powerpoint to covert to PDF...please help

Adobe Reader can't set any security.

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    Dear I forgot my security question answers
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    If you have a rescue email address (which is not the same thing as an alternate email address) set up on your account then steps 1 to 5 on the page that you posted from should let you reset them : go to https://appleid.apple.com/ and click 'Manage your Apple ID' on the right-hand side of that page and log into your account, then click on 'Password and Security' on the left-hand side of that page and on the right-hand side you should see an option to send security question reset info to your rescue email address.
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    or try ringing Apple in your country and ask to talk to the Accounts Security Team : http://support.apple.com/kb/HE57
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  • Still stuck with the same old producer consumer weight problem need help

    Hello All,
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    P1 P2 P3 P4 P5
    5 6 7 8 9
    C1 C2 C3 C4 C5
    2 3 4 5 6
    Now we find all those producer consumer pairs which satisfy the criteria Pi>=Ci
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    (6,6)(7,6)(8,6),(9,6)
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    we assign a third parameter called weight to each element in Si which has satisfied the condition Pi>=Ci;
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    (5,2,ai),(6,2,bi),(7,2,ci)....etc for S1
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    Suppose we get a pair (5,2,3) in S1 and (5,2,3) in S2 we should see that (5,2,3) is not used to compare to compare with any other elements in the same set S2,
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    6 5 4 5 3
    C0 C1 C2 C3 C4
    we have So as (P0,C0) and (P4,C0)
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    S1 =(P0,C1),(P1,C1) and (P4,C1)
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    S2=(P0,C2),(P1,C2) and (P4,C2) so we dont use P0,C2 and P1 and C2 because PO and P1 are already used in S1.
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    Thanks.
    Regards.
    NP
    What I have tried till now.
    I have one more question could you help me with this.
    I have an array list of this form.
    package mypackage1;
    import java.util.*;
    public class DD
    private  int P;
    private  int C;
    private int weight;
    public void set_p(int P1)
    P=P1;
    public void set_c(int C1)
    C=C1;
    public void set_weight(int W1)
    weight=W1;
    public int get_p()
    return P;
    public int get_c()
    return C;
    public int get_x()
    return weight;
    public static void main(String args[])
    ArrayList a=new ArrayList();
    ArrayList min_weights_int=new ArrayList();
    ArrayList rows=new ArrayList();
    ArrayList temp=new ArrayList();
    Hashtable h=new Hashtable();
    String v;
    int o=0;
    DD[] d=new DD[5];
    for(int i=0;i<4;i++)
    d=new DD();
    for(int i=0;i<4;i++)
    d[i].set_p(((int)(StrictMath.random()*10 + 1)));
    d[i].set_c((int)(StrictMath.random()*10 + 1));
    d[i].set_weight(0);
    System.out.println("Producers");
    for(int i=0;i<4;i++)
    System.out.println(d[i].get_p());
    System.out.println("Consumers");
    for(int i=0;i<4;i++)
    System.out.println(d[i].get_c());
    System.out.println("Weights");
    for(int i=0;i<4;i++)
    System.out.println(d[i].get_x());
    for(int i=0;i<4;i++ )
    int bi =d[i].get_c();
    ArrayList row=new ArrayList();
    for(int j=0;j<4;j++)
    if( d[j].get_p() >=bi)
    d[j].set_weight((int)(StrictMath.random()*10 + 1));
    row.add("(" + bi + "," + d[j].get_p() + "," +d[j].get_x() + ")");
    else
    d[j].set_weight(0);
    row.add("null");
    rows.add(row);
    System.out.println(rows);
    int f=0;
    for(Iterator p=rows.iterator();p.hasNext();)
    temp=(ArrayList)p.next();
    String S="S" +f;
    h.put(S,temp);
    String tt=new String();
    for(int j=0;j<4;j++)
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    // System.out.println("In if loop");
    //System.out.println(temp.get(j).toString());
    String l=temp.get(j).toString();
    System.out.println(l);
    //System.out.println("Comma matches" + l.lastIndexOf(","));
    //System.out.println(min_weights);
    f++;
    for(Enumeration e=h.keys();e.hasMoreElements();)
    //System.out.println("I am here");
    int ii=0;
    int smallest=0;
    String key=(String)e.nextElement();
    System.out.println("key=" + key);
    temp=(ArrayList)h.get(key);
    System.out.println("Array List" + temp);
    for( int j=0;j<4;j++)
    String l=(temp.get(j).toString());
    if(l!="null")
    System.out.println("l=" +l);
    [\code]

    In your example you selected the pair with the greatest
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    Also it's not clear to me that there is always a solution,
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    The most obvious approach is to systematically try
    all possibilities until the answer is reached, or there
    are no possibilities left. This means backtracking whenever
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    backtrack one step and try another possibility at this
    step. After all possible choices of the previous step,
    backtrack one more step and so on.
    This seems rather involved, and it probably is.
    Interestingly, if you know Prolog, it is ridiculously
    easy because Prolog does all the backtracking for you.
    In Java, you can implement the algorithm in much the same
    way as Prolog implementations do it--keep a list of all the
    choice points and work through them until success or there
    are none left.
    If you do know Prolog, you could generate lots of random
    problems and see if there is always a solution.

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         FileVault
              Turned off
         Firewall
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    I'd appreciate any ideas anyone has. I'm far from wise about the ins-and-outs of mac security, so I may be missing something. I'm happy to clarify anything.
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    try using different loadvars instances for your send loadvars and for your receive loadvars.

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