Merge pull request #105 from koseokkyu/heapsort

Project Euler solving Problem 01

Project Euler/Problem 01/sol1.c

@@ -0,0 +1,19 @@
+/*
+If we list all the natural numbers below 10 that are multiples of 3 or 5, 
+we get 3, 5, 6 and 9. The sum of these multiples is 23.
+Find the sum of all the multiples of 3 or 5 below 1000.
+*/
+#include <stdio.h>
+
+int main() {
+	int n = 0;
+	int sum = 0;
+	scanf("%d", &n);
+	for (int a = 0; a < n; a++) {
+		if ((a % 3 == 0) || (a % 5 == 0)) {
+			sum += a;
+		}
+	}
+
+	printf("%d\n", sum);
+}

Project Euler/Problem 01/sol2.c

@@ -0,0 +1,24 @@
+/*
+If we list all the natural numbers below 10 that are multiples of 3 or 5,
+we get 3,5,6 and 9. The sum of these multiples is 23.
+Find the sum of all the multiples of 3 or 5 below N.
+'''
+'''
+This solution is based on the pattern that the successive numbers in the series follow: 0+3,+2,+1,+3,+1,+2,+3.
+*/
+#include <stdio.h>
+
+int main() {
+	int n = 0;
+	int sum = 0;
+	scanf("%d", &n);
+
+	int terms = (n - 1) / 3;
+	sum += ((terms)*(6 + (terms - 1) * 3)) / 2; //sum of an A.P.
+	terms = (n - 1) / 5;
+	sum += ((terms)*(10 + (terms - 1) * 5)) / 2;
+	terms = (n - 1) / 15;
+	sum -= ((terms)*(30 + (terms - 1) * 15)) / 2;
+
+	printf("%d\n", sum);
+}

Project Euler/Problem 01/sol3.c

@@ -0,0 +1,49 @@
+/*
+If we list all the natural numbers below 10 that are multiples of 3 or 5,
+we get 3,5,6 and 9. The sum of these multiples is 23.
+Find the sum of all the multiples of 3 or 5 below N.
+'''
+'''
+This solution is based on the pattern that the successive numbers in the series follow: 0+3,+2,+1,+3,+1,+2,+3.
+*/
+#include <stdio.h>
+
+int main() {
+	int n = 0;
+	int sum = 0;
+	int num = 0;
+	scanf("%d", &n);
+	
+	while (1) {
+		num += 3;
+		if (num >= n)
+			break;
+		sum += num;
+		num += 2;
+		if (num >= n)
+			break;
+		sum += num;
+		num += 1;
+		if (num >= n) 
+			break;
+		sum += num;
+		num += 3;
+		if (num >= n) 
+			break;
+		sum += num;
+		num += 1;
+		if (num >= n) 
+			break;
+		sum += num;
+		num += 2;
+		if (num >= n) 
+			break;
+		sum += num;
+		num += 3;
+		if (num >= n) 
+				break;
+		sum += num;
+	}
+
+	printf("%d\n", sum);
+}

Project Euler/Problem 02/so1.c

@@ -0,0 +1,28 @@
+/*
+Problem:
+Each new term in the Fibonacci sequence is generated by adding the previous two terms. By starting with 1 and 2,
+the first 10 terms will be:
+1,2,3,5,8,13,21,34,55,89,..
+By considering the terms in the Fibonacci sequence whose values do not exceed n, find the sum of the even-valued terms.
+e.g. for n=10, we have {2,8}, sum is 10.
+*/
+#include <stdio.h>
+
+int main() {
+	int n = 0;
+	int sum = 0;
+	int i = 1;
+	int j = 2;
+	int temp;
+	scanf("%d", &n);
+
+	while (j <= n) {
+		if ((j & 1) == 0) //can also use(j%2 == 0)
+			sum += j;
+		temp = i;
+		i = j;
+		j = temp + i;
+	}
+
+	printf("%d\n", sum);
+}

Project Euler/Problem 03/sol1.c

@@ -0,0 +1,57 @@
+/*
+Problem:
+The prime factors of 13195 are 5,7,13 and 29. What is the largest prime factor of a given number N?
+e.g. for 10, largest prime factor = 5. For 17, largest prime factor = 17.
+*/
+#include <stdio.h>
+#include <math.h>
+
+int isprime(int no) {
+	int sq;
+
+	if (no == 2) {
+		return 1;
+	}
+	else if (no%2 == 0) {
+		return 0;
+	}
+	sq = ((int)(sqrt(no))) + 1;
+	for (int i = 3; i < sq; i + 2) {
+		if (no%i == 0) {
+			return 0;
+		}
+	}
+	return 1;
+}
+
+int main() {
+	int maxNumber = 0;
+	int n = 0;
+	int n1;
+	scanf("%d", &n);
+	if (isprime(n) == 1)
+		printf("%d", n);
+	else {
+		while (n % 2 == 0) {
+			n = n / 2;
+		}
+		if (isprime(n) == 1) {
+			printf("%d\n", n);
+		}
+		else {
+			n1 = ((int)(sqrt(n))) + 1;
+			for (int i = 3; i < n1; i + 2) {
+				if (n%i == 0) {
+					if (isprime((int)(n / i)) == 1) {
+						maxNumber = n / i;
+						break;
+					}
+					else if (isprime(i) == 1) {
+						maxNumber = i;
+					}
+				}
+			}
+			printf("%d\n", maxNumber);
+		}
+	}
+}

Project Euler/Problem 03/sol2.c

@@ -0,0 +1,23 @@
+/*
+Problem:
+The prime factors of 13195 are 5,7,13 and 29. What is the largest prime factor of a given number N?
+e.g. for 10, largest prime factor = 5. For 17, largest prime factor = 17.
+*/
+#include <stdio.h>
+
+int main() {
+	int n = 0;
+	scanf("%d", &n);
+	int prime = 1;
+	int i = 2;
+	while (i*i <= n) {
+		while (n%i == 0) {
+			prime = i;
+			n /= i;
+		}
+		i += 1;
+	}
+	if (n > 1)
+		prime = n;
+	printf("%d\n", prime);
+}