Two of my friends faced a problem with writing a program which finds the LCM(Lowest Common Multiple)/GCD(Greatest Common Divisor) of two positive integers, so I helped them out by writing two functions for each. I thought many others might be having the same problem. So here are the functions: For LCM: Code: /* a & b are the numbers whose LCM is to be found */ int lcm(int a,int b) { int n; for(n=1;;n++) { if(n%a == 0 && n%b == 0) return n; } } For GCD: Code: /* a & b are the numbers whose GCD is to be found. Given a > b */ int gcd(int a,int b) { int c; while(1) { c = a%b; if(c==0) return b; a = b; b = c; } }
For LCM Code: if(n%a == 0 && n%b == 0) shouldn't this be Code: if(a%n == 0 && b%n == 0) Also just another way to write the loop Code: for(n=1;;n++) { if(n%a == 0 && n%b == 0) return n; } can be Code: for(n=1;a%n == 0 && b%n == 0;n++); return n For GCD Given a > b can be avoided if we have the following code Code: int gcd(int a,int b) { int c; if(a<b) { c = a; a = b; b = c; } while(1) { c = a%b; if(c==0) return b; a = b; b = c; } }
Shabbir, Least Common Multiple(L.C.M.) of 'a' and 'b' is the smallest number 'n' which is both perfectly divisible by 'a' as well as 'b'; i.e. n%a == 0 && n%b == 0.
Right said Satyan, even I was about to post the same. Thanks Shabbir for the modified code for the loop.
This loop won't work, the condition given would prevent the execution even once. The correct code would be: Code: for(n=1;n%a != 0 && n%b != 0;n++); return n
hey guys, Was just interested if with the GCD function if its possible to find the GCD of 3 different numbers rather than just 2? Would be interested to see it. Thanks
hi everyone! I have code following: You see and shift it Code: struct ths { char ht[24]; char tt[21]; int sobd; float m,l,; }; class ts { private: int sots; ths *ts; public: ts() { sots = 0 ; ts = NULL; } void swap(); }; void ts::swap() { int n = sots; for ( int i = 1; i < n; ++i) for (int j = i + 1; j <= n; ++j) if (ts[i].m < ts[j].m) { ths tmp = ts[i]; ts[i] = ts[j]; ts[j] = tg; } } // thanks.
Thanks so much, my friend! this is code in C++, how do you see ? ofcourse, Class in C++ is difficult. What way do you help me to express easy?
hi everybody! Follow you, Can we do in swap 2 structs easier? Code: tmp = ts[i] ; ts[i] = ts[j]; ts[j] = tmp; thanks.
You can swap the struct if they are addresses and not the objects can be done in that way unless they have the overloaded = operator.
