This exercise is taken from www.programmingpraxis.com
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"The sequence of Hamming numbers 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 25, 27, 30, 32, 36, … (A051037) consists of all numbers of the form 2i·3j·5k where i, j and k are non-negative integers. Edsger Dijkstra introduced this sequence to computer science in his book A Discipline of Programming, and it has been a staple of beginning programming courses ever since. Dijkstra wrote a program based on three axioms:
Axiom 1: The value 1 is in the sequence.
Axiom 2: If x is in the sequence, so are 2 * x, 3 * x, and 5 * x.
Axiom 3: The sequence contains no other values other than those that belong to it on account of Axioms 1 and 2.
Your task is to write a program to compute the first thousand terms of the Hamming sequence. When you are finished, you are welcome to read or run a suggested solution, or to post your own solution or discuss the exercise in the comments below."
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Here is my C solution
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"The sequence of Hamming numbers 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 25, 27, 30, 32, 36, … (A051037) consists of all numbers of the form 2i·3j·5k where i, j and k are non-negative integers. Edsger Dijkstra introduced this sequence to computer science in his book A Discipline of Programming, and it has been a staple of beginning programming courses ever since. Dijkstra wrote a program based on three axioms:
Axiom 1: The value 1 is in the sequence.
Axiom 2: If x is in the sequence, so are 2 * x, 3 * x, and 5 * x.
Axiom 3: The sequence contains no other values other than those that belong to it on account of Axioms 1 and 2.
Your task is to write a program to compute the first thousand terms of the Hamming sequence. When you are finished, you are welcome to read or run a suggested solution, or to post your own solution or discuss the exercise in the comments below."
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Here is my C solution
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#include#include int ham_num(int x) { int h = x; while (h && (h%2 == 0)) h = h/2; while (h && (h%3 == 0)) h = h/3; while (h && (h%5 == 0)) h = h/5; /* * h will be 1, when it is ham num, else it will be * something else - example h =10 */ return h; } int main(int argc, char *argv[]) { int x = 1, cnt = 1000; printf("Hamming Numbers : \n"); if (argc > 1) cnt = atoi(argv[1]); while (cnt) { if (ham_num(x) == 1) { printf("%d ",x); cnt--; } x++; } printf("\n"); }