Note: The recursive portions of this code can use a lot of stack space. You may need to tell your linker to allocate more space to prevent a stack overflow exception.
To calculate the factorial of a user given number, simply multiply all of the integers between 2 and that number, inclusive, together. There are two fundamental algorithms; iterative and recursive...
Iterative uses a loop of some sort to iterate through each multiplier and multiply that to the running product.
These examples were developed and checked on the Microsoft Windows SDK 7. Portable compilers might not like the 64-bit long long data type.
/* Microsoft 32-bit iterative */
unsigned long NFactLongIterative (unsigned long N) {
unsigned long result = N;
if (N < 2) return 1;
if (N 2) {
decimal_initialize (d, 2);
return;
}
while (N > 2) {
decimal_multiply (d, N);
N--;
}
return;
}
Main line code, showing full functionality
#include
... all of the above routines are inserted here
/* Example main line */
/* Generates all variations to show differences in results */
int main (int argc, char *argv[]) {
int N;
decimal Decimal = {2, NULL};
if (argc < 2) {
fprintf (stderr, "Usage: factorial N\n");
return 1;
}
N = atoi (argv[1]);
printf ("Long: %u! = %u\n", N, NFactLongIterative (N));
printf ("LongLong: %u! = %I64u\n", N, NFactLongLongIterative (N));
printf ("Recursive: %u! = %I64u\n", N, NFactLongLongRecursive (N));
printf ("Double: %u! = %.0f\n", N, NFactDouble (N));
/* note: arbitrary is exact - if the others don't match, arithmetic overflow occurred */
printf ("Arbitrary: %u! = ", N);
decimal_NFactIterative (&Decimal, N);
decimal_print_digits (&Decimal, TRUE);
return 0;
}
write an algorithm to print the factorial of a given number and then draw the flowchart. This looks like someones homework, defiantly someone looking for the easy way. When it comes to programming, the more you do the better you get. Experience counts (making your own mistakes and learning from the mistake).
double factorial(double N){double total = 1;while (N > 1){total *= N;N--;}return total; // We are returning the value in variable title total//return factorial;}int main(){double myNumber = 0;cout > myNumber;cout
<html> <script language="vbscript"> n=cint(inputbox("Enter a number")) dim f f=1 if n<0 then Msgbox "Invalid number" elseif n=0 or n=1 then MsgBox "The factorial of given number "&n&" is :"&f else for i=n to 2 step -1 f=f*i next MsgBox "The factorial of given number "&n&" is :"&f end if </script> </html>
a write the algorithm to concatenate two given string
Use this function: long factorial(int N){if (N == 0){return 1;}else{return N*factorial(N-1);}}
write an algorithm to print the factorial of a given number and then draw the flowchart. This looks like someones homework, defiantly someone looking for the easy way. When it comes to programming, the more you do the better you get. Experience counts (making your own mistakes and learning from the mistake).
double factorial(double N){double total = 1;while (N > 1){total *= N;N--;}return total; // We are returning the value in variable title total//return factorial;}int main(){double myNumber = 0;cout > myNumber;cout
Factorials are the product of 1 and all the integers up to the given number. Simply put, 5 factorial or 5! = 5*4*3*2*1
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answer:32 programme to print factorial of a given number in c languages
no answer....pls post
The number of diagonals in an n-sided polygon is given by nC2 - n (where n is the number of sides of the polygon) or in the expanded form: factorial (n) _______________________ {factorial (2) * factorial (n-2)} substituting (n = 6) for a hexagon we get the number of diagonals as 9. Similarly, substituting (n=5) for a pentagon we get the number of diagonals as 5.
<html> <script language="vbscript"> n=cint(inputbox("Enter a number")) dim f f=1 if n<0 then Msgbox "Invalid number" elseif n=0 or n=1 then MsgBox "The factorial of given number "&n&" is :"&f else for i=n to 2 step -1 f=f*i next MsgBox "The factorial of given number "&n&" is :"&f end if </script> </html>
Kat
factorial n is given by formula n! = n.(n-1)....1 int i; long x; x =1; for (i=n;i>1;i--) x = x*i ; will calculate factorial. I have put x as long to avoid integer overflow. checks for n is positive etc. to be added.
Complexity of an algorithm is a measure of how long an algorithm would take to complete given