0
#define SIZE 100
#include <stdio.h>
int hcf_function(int,int);
int lcm_function(int,int);
int main()
{
int array[SIZE],n,i,choice,LCM,hcf;
printf("Enter No of Elements\n");
scanf("%d",&n);
printf("Enter Elements\n");
for(i=0;i<n;i++)
scanf("%d",&array[i]);
do
{
printf("\n\nEnter Choice\n\n1.HCF\n2.LCM\n3.Exit\n");
scanf("%d",&choice);
switch(choice)
{
case 1: hcf=array[0];
for(i=1;i<n;i++)
hcf=hcf_function(hcf,array[i]);
printf("\nHCF = %d",hcf);
break;
case 2: LCM=array[0];
for(i=1;i<n;i++)
LCM=lcm_function(LCM,array[i]);
printf("\nLCM = %d",LCM);
break;
case 3: break;
default:printf("Wrong Choice");
break;
}
}while(choice!=3);
}
/***************************************************************
Function Name : hcf_function
Purpose : to find hcf
Input : two numbers
Return Value : hcf
Return Type : int
****************************************************************/
int hcf_function(int m,int n)
{
int temp,reminder;
if(m<n)
{
temp=m;
m=n;
n=temp;
}
while(1)
{
reminder=m%n;
if(reminder==0)
return n;
else
m=n;
n=reminder;
}
}
/***************************************************************
Function Name : lcm_function
Purpose : to find LCM
Input : two numbers
Return Value : LCM
Return Type : int
****************************************************************/
int lcm_function(int m,int n)
{
int LCM;
LCM=m*n/hcf_function(m,n);
return LCM;
}
#include <stdio.h>
#include <stdlib.h>
int getGCF(int val1, int val2);
int main(int argc, char *argv[]){
int val1, val2, gcf;
if(argc != 3){
printf("Syntax: gcf val1 val2\n");
return 1;
}
val1 = atoi(argv[1]);
val2 = atoi(argv[2]);
gcf = getGCF(val1, val2);
if(gcf 1; n--){
if(!(val1 % n | val2 % n)){
rval = n;
}
}
return rval;
}
The recursive method has no loops so the assumption is to use the iterative method instead. Both methods are shown below, but the iterative method uses a goto rather than the usual while loop (commented out).
#include<iostream>
#include<time.h>
//#define RECURSIVE // uncomment to use recursive method
#ifdef RECURSIVE
// Returns the GCD of the two given integers (recursive method)
unsigned int gcd(unsigned int a, unsigned int b)
{
if(!a)
return(b);
if(!b)
return(a);
if(a==b)
return(a);
if(~a&1)
{
if(b&1)
return(gcd(a>>1,b));
else
return(gcd(a>>1,b>>1)<<1);
}
if(~b&1)
return(gcd(a,b>>1));
if(a>b)
return(gcd((a-b)>>1,b));
return(gcd((b-a)>>1,a));
}
#else
// Returns the GCD of the two given integers (iterative method)
unsigned int gcd(unsigned int a,unsigned int b)
{
if(!a)
return(b);
if(!b)
return(a);
int c;
for(c=0; ((a|b)&1)==0; ++c)
{
a>>=1;
b>>=1;
}
while((a&1)==0)
a>>=1;
again:
//do{
while((b&1)==0)
b>>=1;
if(a>b)
{
unsigned int t=a;
a=b;
b=t;
}
b-=a;
//}while(b);
if(b)
goto again;
return(a<<c);
}
#endif
int main()
{
using std::cout;
using std::endl;
srand((unsigned) time(NULL));
for(unsigned int attempt=0; attempt<10; ++attempt)
{
unsigned int x=rand()%100;
unsigned int y=rand()%100;
unsigned int hcf=gcd(x,y);
cout<<"GCD("<<x<<','<<y<<") = "<<hcf<<endl;
}
cout<<endl;
}
#include<iostream>
#include<time.h>
unsigned int gcd(unsigned int a, unsigned int b)
{
if(!a)
return(b);
if(!b)
return(a);
if(a==b)
return(a);
if(~a&1)
{
if(b&1)
return(gcd(a>>1,b));
else
return(gcd(a>>1,b>>1)<<1);
}
if(~b&1)
return(gcd(a,b>>1));
if(a>b)
return(gcd((a-b)>>1,b));
return(gcd((b-a)>>1,a));
}
// Returns the maximum value in the array
unsigned int max(const unsigned int n[],const unsigned int size)
{
unsigned int max=n[0];
for( unsigned int index=1; index<size; ++index )
if(max<n[index])
max=n[index];
return(max);
}
// Returns the minimum non-zero value in the array
unsigned int min_nz(const unsigned int n[],const unsigned int size)
{
unsigned int min=max(n,size);
for( unsigned int index=0; index<size; ++index )
if(n[index]&&n[index]<min)
min=n[index];
return(min);
}
// Returns the greatest common divisor in the array
unsigned int gcd(const unsigned int n[],const unsigned int size)
{
unsigned int hcf=min_nz(n,size);
for( unsigned int index=0; index<size; ++index )
hcf=gcd(hcf,n[index]);
return(hcf);
}
int main()
{
using std::cout;
using std::endl;
srand((unsigned) time(NULL));
for(unsigned int attempt=0; attempt<10; ++attempt)
{
// generate an array of 2 to 4 elements with values in range 0 to 99.
unsigned int size=rand()%3+2;
unsigned int* num = new unsigned int[size];
unsigned int index=0;
while(index<size)
num[index++]=rand()%100;
unsigned int hcf=gcd(num,size);
cout<<"GCD(";
index=0;
cout<<num[index];
while(++index<size)
cout<<','<<num[index];
cout<<") = "<<hcf<<endl;
delete[]num;
}
cout<<endl;
}
#include<stdio.h>
int main()
{
int a,b,c,d1,d2,d3;
int gcd(int ,int);
printf("Enter three integers: ");
scanf("%d%d%d",&a,&b,&c);
d1=gcd(gcd(a,b),c);
printf("Greatest common divisors is: %d",d1);
return 0;
}
int gcd(int x,int y)
{
int c,a,b;
if(x>=y)
{
a=x;
b=y;
}
else
{
a=y;
b=x;
}
while(1)
{
c = a%b;
if(c==0)
return b;
a = b;
b = c;
}
}
$ cat gcd.c
#include
int gcd_recursive(int a, int b)
{
if (b == 0)
return a;
return gcd_recursive(b, a % b);
}
int gcd_nonrecursive(int a, int b)
{
int t;
while (b != 0) {
t = b;
b = a % b;
a = t;
}
return a;
}
int main()
{
printf("GCD of 10 and 25: %d\n", gcd_recursive(10, 25));
printf("GCD of 10 and 25: %d\n", gcd_nonrecursive(10, 25));
return 0;
}
$ gcc gcd.c -o gcd
$ ./gcd
GCD of 10 and 25: 5
GCD of 10 and 25: 5
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Define flowchart and draw flowchart for GCD of two numbers?
pictorial representation of a program is called a flowchart
How do you write a C program to find the GCD of two given integers using non recursive functions?
Use the following function: int gcd (int a, int b) { while (b != 0) { a %= b; a ^= b ^= a ^= b; } return a; } Note that a ^= b ^= a ^= b is an efficient method of swapping two values.
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; }
Write a program to find gcd using recursive method in java?
for two positive integers: public static int gcd(int i1, int i2) { // using Euclid's algorithm int a=i1, b=i2, temp; while (b!=0) { temp=b; b=a%temp; a=temp; } return a; }
Shell program for gcd of three given numbers?
write a shell program for finding out gcd of three given numbers? write a shell program for finding out gcd of three given numbers? write a shell program for finding out gcd of three given numbers? check bellow link http://bashscript.blogspot.com/2009/08/gcd-of-more-than-two-numbers.html
Define flowchart and draw flowchart for GCD of two numbers?
pictorial representation of a program is called a flowchart
Write a c program to find GCD of two given integers by using both recursive n non recursive functions?
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Java program for finding GCD and LCM of two given numbers?
These are the two functions you need: public static int lcm(int i1, int i2) { return (i1*i2/gcd(i1,i2)); } public static int gcd(int i1, int i2) { // using Euclid's algorithm int a=i1, b=i2, temp; while (b!=0) { temp=b; b=a%temp; a=temp; } return a; }
A program to find GCD andLCM of two numbers?
// recursive algorithm to return gcd using Euclid's Algorithm int gcd (int a, int b) { if (a<0) a= -a; if (b<0) b= -b; if (a<b) { int tmp; tmp= a; a= b; b= tmp; } if (b == 0) return a; return gcd (b, a%b); } // LCM using gcd int LCM (int a, int b) { int t; t = a*b; if (t<0) t=-t; return t / gcd (a, b); }
How do you write a C program to find the GCD of two given integers using non recursive functions?
Use the following function: int gcd (int a, int b) { while (b != 0) { a %= b; a ^= b ^= a ^= b; } return a; } Note that a ^= b ^= a ^= b is an efficient method of swapping two values.
How do you find the numbers given the gcd and lcm?
if the gcd and lcm are given and one of the numbers are also given,multiply the gcd and lcm and divide them by the given number
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.
What is the gcd of any two consecutive even numbers?
The GCD is 2.
What is the assembly language program for GCD of two 16 bit unsigned integers using Intel 8086?
alp for lcm of a no
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; }
The greatest common factor of 114 and 190?
The GCD is 38. You can find it with a calculator, using prime factorization with exponents, or using one of the many on-line GCD calculators.You can also list the factors of both numbers and find the largest one that they have in common.
