C Programmer...help(I'll give my 10 points for good answer)?

can you write a program whose output is like this:





How many coins are flipped? 3





HHH


HHT


HTH


HTT


THH


THT


TTH


TTT





the program is all about tossing coin/s (probability), also known as Head and Tail.





it ask the user of how many coins then it prints the outcomes.


the formula for number of outcomes is 2^n (where n is the number of coins).

C Programmer...help(I'll give my 10 points for good answer)?
#include %26lt;math.h%26gt;


void printcorrespondinghandt(int i,int num);


void main()


{


int i;


int num = 3;


for(i=0;i%26lt;pow(2,num);i++)


{


printcorrespondinghandt(i,num);


printf("\n");


}


}





void printcorrespondinghandt(int i,int num)


{


int j;


for(j=0;j%26lt;num;j++)


{


if((i%26gt;%26gt;j)%26amp;0x1)


printf("h");


else


printf("t");


}


}
Reply:umm.. i am not gonna code this for you, but there is an easy way to look at this.


Basically given n coins, you can only get 2^n output.


Assume H=1 and T=0, then it looks like


for(i=0;i%26lt;2^n-1;i++)


print binary representation of i, replacing 1s with H and 0s with T
Reply:#include %26lt;iostream%26gt;


#include %26lt;windows.h%26gt;





using std::cin;


using std::cout;


using std::endl;





inline char* createPattern(char%26amp;, char%26amp;, int%26amp;, int%26amp;);





int main() {





    char head = 'H';


    char tail = 'T';


    int n;





    cout %26lt;%26lt; "#of coins= "; cin %26gt;%26gt; n;





    for (int i = 1 %26lt;%26lt; n; --i %26gt;= 0;) { //1 %26lt;%26lt; n = pow(2, n)


        cout %26lt;%26lt; createPattern(head, tail, i, n) %26lt;%26lt; endl;


    }





    system("PAUSE"); //calls MS-DOS pause command





    return 0;


}








inline char* createPattern(char%26amp; h, char%26amp; t, int%26amp; i, int%26amp; n) {


    char *c = new char[n];


    for (int x = n, j = 0; --x %26gt;= 0;) {


        c[j++] = (i %26amp; (0x1 %26lt;%26lt; x)) == 0 ? t : h;


    }





    return c;


}