Finding GCD of Array Code C language - c

Question

I am trying to write a program in C. The program is supposed to find the GCD (greatest common divisor) of a given array. I am trying to use the smallest number of the array to find the GCD. I was wondering whats wrong with my last loop. I havent figured a way on how to check if the division is giving any decimal points in order to stop the loop. This is my code
#include <stdio.h>
#include <stdlib.h>
int main()
{
int A[10]={112, 160, 180, 240, 288, 32, 480, 96, 60, 72};
int i;
int j;
int minimum = A[0];
int GCD;
int temp;
for (i=1;i<9;i++)
{
if( A[i] < minimum)
{
minimum = A[i];
}
}
for (i=1; i < minimum/2; i++)
{
for (j = 0; j < 9;j++)
{
GCD = 2*i;
temp = ((A[j])/(GCD));
int check = temp%1;
if (check == 0)
break;
}
}
printf("The Greates Common Denominator is: %d", GCD);
return 0;
}

Answer 1

#include <stdio.h>
unsigned gcd(unsigned x, unsigned y){
unsigned wk;
if(x<y){ wk=x;x=y;y=wk; }
while(y){
wk = x%y;
x=y;
y=wk;
}
return x;
}
int gcd_a(int n, int a[n]){
if(n==1) return a[0];
if(n==2) return gcd(a[0], a[1]);
int h = n / 2;
return gcd(gcd_a(h, &a[0]), gcd_a(n - h, &a[h]));
}
int main(void){
int A[10]={112, 160, 180, 240, 288, 32, 480, 96, 60, 72};
int size_A = sizeof(A)/sizeof(*A);
int gcd = gcd_a(size_A, A);
printf("%d\n", gcd);
return 0;
}

Answer 2

#include <stdio.h>
#include <stdlib.h>
int gcd(int a, int b)
{
int r;
while(a)
{
r=b%a;
b=a;
a=r;
}
return b;
}
int gcd_(int* A, int N)
{
int c=gcd(*A,*(A+1));
int i,g;
for(i=1;i<N-1;i++)
{
g=gcd(c,*(A+1+i));
c=g;
}
return c;
}
int main(void)
{
int A[]={96,60,32,72,84,90};
int N=sizeof(A)/sizeof(A[0]);
int t=gcd_(A,N);
printf("%d",t);
return 0;
}

Answer 3

#include <stdio.h>
static int gcd(int x, int y)
{
int r;
if (x <= 0 || y <= 0)
return(0);
while ((r = x % y) != 0)
{
x = y;
y = r;
}
return(y);
}
int main(void)
{
int A[10] = { 112, 160, 180, 240, 288, 32, 480, 96, 60, 72 };
int g = A[0];
for (int i = 1; i < 10; i++)
g = gcd(g, A[i]);
printf("The Greatest Common Denominator is: %d\n", g);
return 0;
}
The answer is 4. This is clearly correct; it is the GCD of 32 and 60; everything else is divisible by 4.
If desired, you could optimize the loop with:
for (int i = 1; i < 10 && g != 1; i++)
g = gcd(g, A[i]);
When the GCD is 1, it won't get any bigger,

Answer 4

#include <iostream>
using namespace std;
int gcd(int a, int b){
int t;
while(a)
{
t = a;
a = b%a;
b = t;
}
return b;
}
int main(){
int n;
cout<<"how many numbers (Max 10): "; // your choice
cin>>n;
cin.ignore();
cout<<endl;
int arr[n-1];
for (int i = 0; i < n; ++i) {
cin>>arr[i];
}
int ans;
ans = arr[0];
for (int j = 0; j < n; ++j) {
ans = gcd(ans,arr[j]);
}
cout<<"GCD = "<<ans;
//cin.get();
return 0;
}

Answer 5

#include <stdio.h>
#include <stdlib.h>
int main() {
int i, j, n, flag = 0, small, *data;
/* get the number of inputs from the user */
printf("Enter the number of inputs:");
scanf("%d", &n);
/* allocate memory to store n numbers */
data = (int *)malloc(sizeof(int) * n);
/* get n numbers from the user */
for (i = 0; i < n; i++) {
printf("Data[%d]: ", i);
scanf("%d", &data[i]);
}
/* find the smallest of n numbers */
small = data[0];
for (i = 1; i < n; i++) {
if (data[i] < small)
small = data[i];
}
/*
* use the smallest no to find gcd of n numbers.
* Start checking from small to 1 whether the
* same value divides all the given inputs
*/
for (i = small; i > 0; i--) {
for (j = 0; j < n; j++) {
if (data[j] % i != 0) {
flag = 1;
}
}
/* print the result */
if (!flag) {
printf("GCD of given %d numbers is %d\n", n, i);
break;
}
flag = 0;
}
return 0;
}

Answer 6

#include <stdio.h>
#include <conio.h>
int x;
void main()
{
int i,num[10]={0};
printf("How many numbers do you want to enter =");
scanf("%d",&x);
for(i=0;i<x;i++)
scanf("%d",& num[i]);
printf("The hcf of the numbers are = %d ", hcf(num));
}
int hcf(int num[])
{
int count,rem,lv,i,j;
lv=gnum(num);
if(lv==1)
return 1;
for(j=2;j<=lv;j++)
{
count =0;
for(i=0;i<x;i++)
{
rem = num[i]%j ;
if( rem != 0 || num[i]< j )
break;
else
{
count++;
}
}
if(count == x)
{
for(i=0;i<x;i++)
num[i]=num[i]/j;
return(j*hcf(num));
}
}
if(count!= x)
return 1;
}
int gnum(int num[])
{
int i,temp=num[0];
for(i=0;i<x;i++)
{
if(temp >= num[i])
temp = num[i];
}
return (temp);
}

Answer 7

#include <stdio.h>
#include<stdlib.h>
int main(void)
{
int A[2] = {10,30};
int a,b,r,i;
for ( i = 1; i < 2; i++)
{
a=A[0],b=A[i];
while(b!=0)
{
r=a%b;/*remainder*/
a=b;
b=r;
}
}
printf("The Greatest Common Denominator is: %d\n", a);
return 0;
}
/*You can change the **for** loop range according to the array values*\

Answer 8

#include <stdio.h>
#include <stdlib.h>
int main()
{
int A[10]={112, 160, 180, 240, 288, 32, 480, 96, 60, 72};
int i;
int j;
int minimum = A[0];
int GCD;
int temp;
int f;
for (i=1;i<10;i++)
{
if( A[i] < minimum)
{
minimum = A[i];
}
}
for (i=1; i <= minimum; i++)
{
f=0;
for (j = 0; j < 10;j++)
{
if(A[j]%i!=0)
{
f=1;
break;
}
}
if(f==0)
GCD=i;
}
printf("The Greates Common Denominator is: %d ", GCD);
return 0;
}
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