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authorNguyễn Gia Phong <vn.mcsinyx@gmail.com>2019-12-28 21:16:58 +0700
committerNguyễn Gia Phong <vn.mcsinyx@gmail.com>2019-12-28 21:21:44 +0700
commit0887d8f96950a060a122e14ed2981182ff1eb37d (patch)
tree68be7a59c323c1fd901455ffae8f21874c4bb4c6 /usth/ICT2.1/final/problem1.c
parente461df7573c2b7b7e26c965d8cf2d8e175d67378 (diff)
downloadcp-0887d8f96950a060a122e14ed2981182ff1eb37d.tar.gz
[usth/ICT2.1] Algorithm and Data Structures
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+/* Read a natural number from stdin and print its divisors to stdout */
+#include <stdio.h>
+
+/* Print all divisors of the given natural number n. */
+void print_divisors(int n, int i)
+{
+	if (i * i == n)
+		printf("%d\n", i);		/* Θ(1) */
+	if (i * i >= n)
+		return;				/* Θ(1) */
+	if (n % i == 0)
+		printf("%d\n%d\n", i, n / i);	/* Θ(1) */
+	print_divisors(n, i + 1);
+}
+
+/*
+ * Complexity analysis:
+ * Let T(n, i) be the time complexity of print_divisors, it is obvious that
+ *     T(n, i) = Θ(1) if i*i >= n
+ *     T(n, i) = T(n, i + 1) + Θ(1)
+ * Thus T(n, 1) = sum(Θ(1) for i from 1 to ceil(sqrt(n))) = Θ(sqrt(n))
+ * or the overall time complexity of the call from main is Θ(sqrt(n))
+ * on all scenarios.
+ */
+int main()
+{
+	int n;
+
+	scanf("%d", &n);
+	print_divisors(n, 1);
+
+	return 0;
+}