Programming
Aptitude
Syllabus
Interview Preparation
Interview Exp.
Off Campus
Programming
Aptitude
Syllabus
Interview Preparation
Interview Exp.
Off Campus
Programming
Aptitude
Syllabus
Interview Preparation
Interview Exp.
Off Campus
0
#include
int main()
{
int i,l,m,x,y,z;
printf(“enter 3 numbers”);
scanf(“%d%d%d”,&x,&y,&z);
if(x<y)
{
if(x<z)
m=x;
else
m=z;
}
else
if(y<z)
m=y;
else
m=z;
for(i=1;i<=m;++i)
{
if(x%i==0&&z%i==0&&z%i==0)
l=i;
}
printf("\n%d",l);
return 0;
}
#python3
def gcd(nums):
lis=[]
for i in nums:
final=[]
for j in range(1,i+1):
if i%j==0:
final.append(j)
else:
continue
lis.append(final)
finals=[set(x) for x in lis[1:]]
return ‘the gcd of the given terms is {}’.format(max(list(set(lis[0]).intersection(*finals))))
list1=[]
n=int(input(‘enter the number of terms:\t’))
for i in range(n):
inp=int(input(‘enter the number:\t’))
list1.append(inp)
print(gcd(list1))
#include
using namespace std;
int gcd(int a,int b) {
int rem;
if(a > b) {
int temp = b;
b = a;
a = temp;
}
while(a) {
rem = b%a;
b = a;
a = rem;
}
return b;
}
int main()
{
int a,b,c,ans;
cin>>a>>b>>c;
ans = gcd(gcd(a,b),c);
cout<<ans;
return 0;
}
#include
int main()
{
int a,b,c,hcf;
scanf(“%d %d %d”,&a,&b,&c);
hcf=(a<b)?(a<c)?a:c:(b<c)?b:c;
while(1)
{
if(a%hcf==0 && b%hcf==0 && c%hcf==0)
break;
hcf–;
}
printf("%d",hcf);
return 0;
}
Thanks for contributing your code Sadique
#python Program
from fractions import gcd
def LCMofArray(a):
lcm = a[0]
for i in range(1,len(a)):
lcm = gcd(lcm, a[i])
return lcm
arr1 = [120,60,40,20]
print(“LCM of arr1 elements:”, LCMofArray(arr1))
Thanks for contributing the code Bathula
#python
from fractions import gcd
def LCMofArray(a):
lcm = a[0]
for i in range(1,len(a)):
lcm = gcd(lcm, a[i])
return lcm
arr1 = [120,60,40,20]
print(“LCM of arr1 elements:”, LCMofArray(arr1))
Thanks for contributing the code Bathula
SOLUTION IN PYTHON
l = []
n = int(input(‘Enter the size of array of which you want to find GCD: ‘))
for i in range(n):
x = int(input())
l.append(x)
x = l[0]
y = l[1]
def GCD(x,y):
if x<y:
x, y = y, x
while y!=0:
x,y = y, x%y
return x
gcd = GCD(x,y) #finding gcd of first 2 numbers
for i in range(2, len(l)):
gcd = GCD(gcd, l[i])
print(gcd)
Thanks for contributing the code Tirtha
n1,n2,n3=map(int,input().split(“,”))
if(n1<n2 and n2<n3 and n1<n3):
x=n1
elif(n1n3 and n1n2 and n2<n3 and n1<n3):
x=n2
elif(n1n3 and n1>n3):
x=n3
elif(n1>n2 and n2n3):
x=n2
elif(n1>n2 and n2>n3 and n1>n3):
x=n3
for i in range(1,x+1):
if(n1%i==0 and n2%i==0 and n3%i==0):
gcd=i
print(f”GCD for {n1},{n2},{n3} is:{gcd}”)
#python
n1,n2,n3=map(int,input().split(“,”))
if(n1<n2 and n2<n3 and n1<n3):
x=n1
elif(n1n3 and n1n2 and n2<n3 and n1<n3):
x=n2
elif(n1n3 and n1>n3):
x=n3
elif(n1>n2 and n2n3):
x=n2
elif(n1>n2 and n2>n3 and n1>n3):
x=n3
for i in range(1,x+1):
if(n1%i==0 and n2%i==0 and n3%i==0):
gcd=i
print(f”GCD for {n1},{n2},{n3} is:{gcd}”)
#include
int main()
{
int n1, n2,n3,i, gcd;
printf(“Enter two integers: “);
scanf(“%d %d %d”, &n1, &n2,&n3);
for(i=1; i <= n1 && i <= n2 && i<=n3; ++i)
{
// Checks if i is factor of both integers
if(n1%i==0 && n2%i==0 && n3%i==0)
gcd = i;
}
printf("G.C.D of %d and %d is %d %d", n1, n2,n3,gcd);
return 0;
}
,n3,i, gcd;
printf(“Enter two integers: “);
scanf(“%d %d %d”, &n1, &n2,&n3);
for(i=1; i <= n1 && i <= n2 && i<=n3; ++i)
{
// Checks if i is factor of both integers
if(n1%i==0 && n2%i==0 && n3%i==0)
gcd = i;
}
printf("G.C.D of %d and %d is %d %d", n1, n2,n3,gcd);
return 0;
}
its simple way to gets GCD……//////
#include
int main()
{
int a,b,c, gcd;
scanf(“%d %d %d”,&a,&b,&c);
for(int i=1; i <=a && i <=b && i<=c; ++i)
{
// Checks if i is factor of both integers
if(a%i==0 && b%i==0 && c%i==0)
gcd = i;
}
printf("G.C.D is %d",gcd);
return 0;
}