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Determination of Prime Factors using Functions.
A positive integer is entered through the keyboard.
Write a function to obtain the prime factors of this number.
For example, prime factors of 24 are 2, 2, 2 and 3, whereas prime factors of 35 are 5 and 7.



#include
main()
{

int number;
int prime(int number);

int primefactor(int number);


printf("Enter the number whose prime factors are to be calculated:");

scanf ("%d", &number);

primefactor(number);

}

//The following function detects a Prime number.

prime(int num)
{
int i, ifprime;


for (i=2; i<=num-1; i++)

{
if (num%i==0)
{

ifprime=0;
}
else
ifprime=1;

}
return (ifprime);
}


//The following function prints the prime factors of a number.

primefactor(int num)
{
int factor,ifprime;
for (factor=2; factor<=num;)

{
prime(factor); //so that the factors are only prime and nothing else.
if (ifprime)

{
if (num%factor==0) //diving by all the prime numbers less than the number itself.
{

printf("%d ", factor);
num=num/factor;

continue;
}
else
{
factor++;//this cannot be made a part of the for loop

}
}
}
return 0;
}





Calculation of A to the power of B using Functions
Write a function power(a,b), to calculate the value of a raised to b.



#include
main()
{

int power (a,b);
int a, b, result;

printf("Enter the value of a and b:");
scanf ("%d %d", &a, &b);

result=power(a,b);
printf("%d raised to %d is %d", a, b, result);

}

power (int a, int b)
{

int calculation=1, calc;
for (calc=1; calc <=b; calc++)

{
calculation=calculation*a;
continue;
}

return(calculation);
}



Conver the given year to Roman Numerals using Functions.
Write a general-purpose function to convert any given year into its roman equivalent.
The following table shows the roman equivalents of decimal numbers:

Decimal:........Roman
1.....................i
5....................v
10..................x
50..................l
100................c
500...............d
1000.............m

Example:
Roman equivalent of 1988 is mdcccclxxxviii
Roman equivalent of 1525 is mdxxv

This program is a big lengthy owing to the use of Case Statements. This program can also be rewritten using Arrays, which will reduce the length considerably.


#include
main()
{

int year;
int convert (int year);



{

printf("Note:Enter a four year digit year.\n\n");

printf("Enter the year that you wanna convert to Roman: " );

scanf ("%d", &year);

if (year> 1999)

{
printf("Invalid Year.Please enter again.\n\n");
}
}

convert(year);


}



convert(int year)

{
int i;

printf("\nYear converted to Roman:");


i=(year/1000); //thousands place
if(i==1)

{
printf("m");
}


i=((year/100)%10); //hundreds place

switch (i)
{
case 1:
printf("c");

break;

case 2:
printf("cc");

break;

case 3:
printf("ccc");

break;

case 4:
printf("cd");

break;

case 5:
printf("d");

break;

case 6:
printf("dc");

break;

case 7:
printf("dcc");

break;

case 8:
printf("dccc");

break;

case 9:
printf("dcccc"); //this part you may think is wrong..9 -> cm

break; //but i have taken a hint from the example in the question.

}



i=((year/10)%10); //tens place

switch(i)
{
case 1:
printf("x");

break;

case 2:
printf("xx");

break;

case 3:
printf("xxx");

break;

case 4:
printf("xl");

break;

case 5:
printf("l");

break;

case 6:
printf("lx");

break;

case 7:
printf("lxx");

break;

case 8:
printf("lxxx");

break;

case 9:
printf("lxxxx"); //had it not been for this example, it would have been xc

break;

}



i=year%10; //ones place

switch(i)
{
case 1:
printf("i");

break;

case 2:
printf("ii");

break;

case 3:
printf("iii");

break;

case 4:
printf("iv");

break;

case 5:
printf("v");

break;

case 6:
printf("vi");

break;

case 7:
printf("vii");

break;

case 8:
printf("viii");

break;

case 9:
printf("ix");

break;
}


printf ("\n\n");

return 0;

}







Detection of Leap year using Functions.
Any year is entered through the keyboard.
Write a function to determine whether the year is a leap year or not.



#include
main()
{

int leap_year(year);
int year, lp;

printf("Enter the year:");
scanf ("%d", &year);

lp=leap_year(year);

if (lp)

{
printf("\nThe entered year is a leap year.");
}
else
{

printf("\nThe entered year is not a leap year.");
}

}


leap_year(int y)

{
int lp;

if (y%4==0)

{
lp=1;
}
else
lp=0;

return(lp);
}