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#include<stdio.h>
#include<conio.h>
int find_gcd(int,int);
int find_lcm(int,int);
int main(){
int num1,num2,gcd,lcm;
clrscr();
printf("\nEnter two numbers:\n ");
scanf("%d %d",&num1,&num2);
gcd=find_gcd(num1,num2);
printf("\n\nGCD of %d and %d is: %d\n\n",num1,num2,gcd);
if(num1>num2)
lcm = find_lcm(num1,num2);
else
lcm = find_lcm(num2,num1);
printf("\n\nLCM of %d and %d is: %d\n\n",num1,num2,lcm);
return 0;
}
int find_gcd(int n1,int n2){
while(n1!=n2){
if(n1>n2)
return find_gcd(n1-n2,n2);
else
return find_gcd(n1,n2-n1);
}
return x;
}
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Judy
Beau
Blake#include<stdio.h>
#include<conio.h>
int GCD (int a,int b)
{
if (a<0) a= -a;
if (b<0) b= -b;
if (a==0 b==1 a==b) return b;
if (a==1 b==0) return a;
if (a>b) return GCD (b, a%b);
else return GCD (a, b%a);
}
void main()
{
int x,y;
clrscr();
printf("nEnter 1st number:");
scanf("%d",&x);
printf("nEnter 2nd number:");
scanf("%d",&y);
printf("\nGCD is:%d", GCD(x,y));
getch();
}
And for three number:
int GCD3 (int x,int y, int z)
{
return GCD(x,GCD(y,z));
}
http://wiki.answers.com/Q/Write_a_recursive_algorithm_to_find_the_gcd_of_two_numbers GCD algorithm....
GCD(BIG,SMALL)
BIG=SMALL*INTEGER + REMAINDER
SMALL=REMAINDER*INTEGER + REM#2
REMAINDER=REM#2*INTEGER+REM#3
When the remainder is 0, then the answer is what's being multiplied by an integer.
Example
GCD(1071,1029)
1071=1029*1 + 42
1029=42*24 + 21
42=21*2+0
GCD=21
Niraj Sharma
There are two methods which are generally used:
- Euclid's method
- Divide the larger number by the smaller to give a result and remainder.
- If the remainder is zero (0), the smaller number is the GCD.
- Replace the larger by the smaller
- Replace the smaller by the remainder
- Repeat the process.
4158 ÷ 1260 = 3 r 378
1260 ÷ 378 = 3 r 126
378 ÷ 126 = 3 r 0
GCD of 1260 and 4158 is 126.
- Method of prime factorisation.
1260 = 22 x 32 x 5 x 7
4158 = 2 x 33 x 7 x 11
GCD of 1260 and 4158 is 2 x 32 x 7 = 126
The prime factorisation method is useful in that it can also be used to find the LCM of the numbers by multiplying together the primes across both the numbers to their highest power, example LCM of 1260 and 4158 is 22 x 33 x 5 x 7 x 11 = 41580.
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Define flowchart and draw flowchart for GCD of two numbers?
pictorial representation of a program is called a flowchart
What is pseudo code for GCD of two numbers?
public class GCD { public static void main(String[] args) { //Example how to use this method System.out.println(GCD(15,50)); } //find the greatest common divisor of two numbers public static int GCD(int a, int b){ if (b == 0) return a; return GCD(b, a % b); } } Hope this help to solve you problem.
How do you write a C program to find the GCD and LCM of two numbers using a switch statement?
The following function will return the GCD or LCM of two arguments (x and y) depending on the value of the fct argument (GCD or LCM). enum FUNC {GCD, LCM}; int gcd_or_lcm(FUNC fct, int x, int y) { int result = 0; switch (fct) { case (GCD): result = gcd (x, y); break; case (LCM): result = lcm (x, y); break; } return result; }
How do you write a algorithm that gives the GCD of two given numbers?
algorithm GCD (a, b) is:while (a b) doif a > b then a := a - b else b := b - aend whilereturn a
Write a C program to find GCD of 2 numbers using non-recursion?
One way to find the GCD (Greatest Common Divisor) of two numbers is Euclid's method. The following program demostrates this, without using recursion. The third number printed is the GCD of the first two. The highlighted lines are the core of the algorithm.#include int GcdByEuclid (int a, int b) {if (a < 0 b < 0) return -1;while (a > 0 && b > 0) if (a > b) a -= b; else b -= a;if (a == 0) return b; else return a;}int main (int argc, char *argv[]) {int a, b;if (argc < 3) {fprintf (stderr, "Usage: gcd a b\n");return 1;}a = atoi(argv[1]);b = atoi(argv[2]);printf ("%d %d %d", a, b, GcdByEuclid (a, b));return 0;}
