It's no problem at all. I actually used one of the previous posters' code and fixed it out to make mine, all it needed was a little push. At first I thought my code wouldn't work both ways and made one similar to yours too, but it ended up working out so I kept and just posted it for future reference for anyone else. I am done with this class (finished with about 99.4%) so I might not be posting for a long time until I take another programming class. Have fun coding!
Here's another way to do this depending on what exactly you are looking for. I can't post the link for some reason which is really annoying . Code: void main() { int n1,n2; clrscr(); printf("\nEnter two numbers:"); scanf("%d %d",&n1,&n2); while(n1!=n2) { if(n1>=n2-1) n1=n1-n2; else n2=n2-n1; } printf("\nGCD=%d",n1); getch(); } Logic of HCF or GCD of any two numbers: In HCF we try to find any largest number which can divide both the number. For example: HCF or GCD of 20 and 30 Both number 20 and 30 are divisible by 1, 2,5,10. HCF=max (1, 2, 3, 4, 10) =10 Logic for writing program: It is clear that any number is not divisible by more than that number. In case of more than one number s, a possible maximum number which can divide all of the number s must be minimum of all of that numbers. For example: 10, 20, and 30 Min (10, 20, 30) =10 can divide all there numbers. So we will take one for loop which will start form min of the numbers and will stop the loop when it became one, since all numbers are divisible by one. Inside for loop we will write one if conditions which will check divisibility of both the numbers. Program: void main(){ int x,y,m,i; clrscr(); printf("Insert any two number: "); scanf("%d%d",&x,&y); m=x for(i=m;i>=1;i--){ if(x%i==0&&y%i==0){ printf("\nHCF of two number is : %d",i) ; break; } } getch(); }
You can't post links cos you're new. It's a measure designed to stop spammers creating accounts and spamming us with links. This restriction is lifted after you've posted a few times and given us chance to see you're someone we want around.
Good code sample. But for() loop should be this way; Code: for(n=1;n%a != 0 || n%b != 0;n++); return n;
For finding the GCD of two positive integers, you can simply write: Code: int gcd(int a,int b) { while((a=a%b)&&(b=b%a)); return a+b; }
OMG!!! You guys are got to be kidding right??? I just went through replies and 99% of them don’t make any sense. I just created the account to reply specifically to this post. COKEDUDE your last reply to this post is nonsense because it doesn't calculate the LCM. But however the answer is a CM. For example let's take number 18 and 24 and you want to find GCD(24,18). There are 2 methods we can use to find GCD. Those are, 1) Prime Factorization According to the Prime Factorization you factorize the each number and find the "overlap" of the two expressions like following. 18 = 2*3*3 24 = 2*2*2*3 Therefore the overlap value is 2*3 Therefore GCD(18,24) = 2*3 = 6 2) Euclid's Algorithm This is much more efficient method. Divide 24 by 18 to get a quotient of 1 and a remainder of 6. Then divide 12 by 6 to get a quotient of 2 and a remainder of 0. Then divide 6 by 0 to get a remainder of 6, which means that 6 is the GCD. Therefore GCD(18,24) = 2*3 = 6 Bellow is a code I wrote which will illustrate the GCD in both methods. Code: #include <stdio.h> #include <conio.h> int PrimeFactorization(int num1, int num2); int Euclid(int num1, int num2); void main(){ int num1 = 0, num2 = 0, primeFactorization = 0, euclid=0; printf("\nEnter the 1st Integer: "); scanf("%d", &num1); printf("Enter the 2nd Integer: "); scanf("%d", &num2); primeFactorization = PrimeFactorization(num1, num2); printf("\n\n\tThe GCD of %d and %d is (Prime Factorization): %d",num1,num2,primeFactorization); euclid = Euclid(num1, num2); printf("\n\n\tThe GCD of %d and %d is (Euclid's Algorithm): %d",num1,num2,euclid); } int PrimeFactorization(int num1, int num2){ int temp =1, i = 2,min = 0; if (num1>num2) min = num2; else min = num1; while(i<min) { if((num1%i == 0) && (num2%i == 0)) { num1 /= i; num2 /= i; temp *= i; } else i++; } return temp; } int Euclid(int num1, int num2){ int temp =1, i = 2,min = 0,max = 0,remainder = 0; if (num1>num2){ min = num2; max = num1; } else{ min = num1; max = num2; } while(min >= remainder){ remainder = max % min; //printf("\n\tmin: %d\tmax: %d\t remainder: %d\t\n",min,max,remainder); if (remainder == 0){ break; } else{ max = min; min = remainder; } } return min; }
Infact to compute the LCM of multple numbers say a, b, c, d quickly is: Code: int LCM(int a, int b, int c, int d) { int max = a; if(max < b) max = b; if(max < c) max = c; if(max < d) max = d; int lcm = max; while(1) { if(lcm%a == 0 && lcm%b == 0 && lcm%c == 0 && lcm%d == 0) return lcm; lcm += max; } }
Re: Finding LCM & GCD in C++ To get full source code for making a program for lcm and gcd calculation in c++, visit this cool link: bitsbyta.blogspot
Correction: yes there are. Direct links: http://bitsbyta.blogspot.com/2011/01/least-common-multiple-lcm-c.html http://bitsbyta.blogspot.com/2011/01/greatest-common-divisor-gcd-c.html
