init p1 to p9

p1.c

@@ -0,0 +1,24 @@
+
+/*
+ * http://projecteuler.net
+ * 
+ * Projecteuler - Problem 1
+ * ------------------------
+ * 2008-01-21 Henrik Hautakoski
+ */
+
+#include <stdio.h>
+
+#define mod(x) ((x % 5) == 0 || (x % 3) == 0)
+
+int main() {
+	
+	int i, r = 0;
+	
+	for(i=3; i < 1000; i++)
+		if (mod(i)) r += i;
+	
+	printf("%i\n", r);
+	
+	return 0;
+}

p2.c

@@ -0,0 +1,29 @@
+
+/*
+ * http://projecteuler.net
+ * 
+ * Projecteuler - Problem 2
+ * -----------------------
+ * 2008-03-11 Henrik Hautakoski
+ */
+
+#include <stdio.h>
+
+int main() {
+	
+	int c = 0, a = 1, b = 1, r = 0;
+		
+	while(c < 4000000) {
+		
+		c = a+b;
+		a = b;
+		b = c;
+			
+		if((c % 2) == 0)
+			r += c;
+	}
+		
+	printf("%i\n", r);
+
+	return 0;
+}

p3.c

@@ -0,0 +1,44 @@
+
+/*
+ * http://projecteuler.net
+ * 
+ * Projecteuler - Problem 3
+ * ------------------------
+ * 10/01/10 (binary day :D) Henrik Hautakoski
+ * 
+ * fast solution using euclidean algorithm for calculating gcd.
+ */
+
+#include <stdio.h>
+
+typedef unsigned long bint;
+
+bint gcd(bint a, bint b) {
+	
+	bint c;
+	
+	while(b != 0) {
+		c = b;
+		b = a % b;
+		a = c;
+	}
+	
+	return a;
+}
+
+int main() {
+	
+	bint i, r = 0, n = 600851475143;
+	
+	for(i=2; i <= n ; i++) {
+		
+		if((r = gcd(n, i)) == 1)
+			continue;
+		
+		n /= r;
+	}
+	
+	printf("%li\n", r);
+	
+	return 0;
+}

p4.c

@@ -0,0 +1,58 @@
+
+/*
+ * http://projecteuler.net
+ * 
+ * Projecteuler - Problem 4
+ * ------------------------
+ * 2009-12-28 Henrik Hautakoski
+ * 
+ * A somewhat elegant solution (representing the numbers as a string is not so fancy)
+ */
+
+#include <stdio.h>
+
+int factor(int val) {
+	
+	int a;
+	
+	for(a=999; a > 99; a--) {
+		
+		if((val % a) == 0 && (val/a) < 999)
+			return a;
+	}
+	
+	return 0;
+}
+
+int ispalindrom(int num) {
+	
+	int i, n;
+	char buffer[7];
+	
+	sprintf(buffer, "%i", num);
+	for(n=0; buffer[n+1] != 0; n++);
+	
+	for(i=0; i < n-i; i++) {
+		
+		if (buffer[i] != buffer[n-i])
+			return 0;
+	}
+	
+	return 1;
+}
+
+int main() {
+	
+	int r, a;
+	
+	for(r=999999; r >= 100001; r--) {
+		
+		if (!ispalindrom(r) || !(a = factor(r)))
+			continue;
+		
+		printf("%i * %i = %i\n", r/a, a, r);
+		break;
+	}
+	
+	return 0;
+}

p5.c

@@ -0,0 +1,31 @@
+
+/*
+ * http://projecteuler.net
+ * 
+ * Projecteuler - Problem 5
+ * ------------------------
+ * 2008-03-11 Henrik Hautakoski
+ */
+
+#include <stdio.h>
+
+int main() {
+    
+    int i, n = 20;
+    char r;
+    
+    do {
+        r = 0;
+        for(i=20; i > 10; i--) {
+            if ((n % i) != 0) {
+                n += 20;
+                r = 1;
+                break;
+            }
+        }
+    } while(r);
+    
+    printf("%i\n", n);
+    
+    return 0;
+}

p6.c

@@ -0,0 +1,22 @@
+
+/* 
+ * http://projecteuler.net
+ * 
+ * Projecteuler - Problem 6
+ * ------------------------
+ * 2008-03-13 Henrik Hautakoski
+ */
+
+#include <stdio.h>
+
+int main() {
+    
+    int i, s1 = 1, s2 = 1;
+    
+    for(i=2; i <= 100; i++) {
+        s1 += i*i;
+        s2 += i;
+    }
+    
+    printf("%i\n", s2*s2 - s1);
+}

p7.c

@@ -0,0 +1,55 @@
+
+/*
+ * http://projecteuler.net
+ * 
+ * Projecteuler - Problem 7
+ * ------------------------
+ * 2009-12-28 Henrik Hautakoski
+ * 
+ * fast and elegant solution using The fundamental theorem of arithmetic wich state: 
+ * any natural number greater than 1 can be prime factorized in only one unique way 
+ * unless it is prime itself.
+ */
+
+#include <stdio.h>
+#include <malloc.h>
+
+int isprime(int *pl, int l, int p) {
+    
+    int i;	
+	
+	for(i=0; i < l; i++) {
+
+		if((p % pl[i]) == 0)
+			return 0;
+	}
+	
+	return 1;
+}
+
+int main() {
+	
+    int i, len = 3, *prime = malloc(sizeof(int)*10001);
+	
+    if (prime == NULL)
+        return 1;
+    
+    prime[0] = 2;
+    prime[1] = 3;
+    prime[2] = 5;
+    
+	for(i=6; len < 10001; i++) {
+		
+		if(!isprime(prime, len, i))
+			continue;
+		
+		prime[len++] = i;
+	}
+
+	printf("%i\n", prime[len-1]);
+    
+    free(prime);
+    
+	return 0;
+}
+

p8.c

@@ -0,0 +1,65 @@
+
+/*
+ * http://projecteuler.net
+ * 
+ * Projecteuler - Problem 8
+ * ------------------------
+ * 2009-12-28 Henrik Hautakoski
+ * 
+ * No comment!
+ */
+
+#include <stdio.h>
+
+#define N_CHARS 1000
+
+#define BIGWALLOFTEXT \
+"73167176531330624919225119674426574742355349194934\
+96983520312774506326239578318016984801869478851843\
+85861560789112949495459501737958331952853208805511\
+12540698747158523863050715693290963295227443043557\
+66896648950445244523161731856403098711121722383113\
+62229893423380308135336276614282806444486645238749\
+30358907296290491560440772390713810515859307960866\
+70172427121883998797908792274921901699720888093776\
+65727333001053367881220235421809751254540594752243\
+52584907711670556013604839586446706324415722155397\
+53697817977846174064955149290862569321978468622482\
+83972241375657056057490261407972968652414535100474\
+82166370484403199890008895243450658541227588666881\
+16427171479924442928230863465674813919123162824586\
+17866458359124566529476545682848912883142607690042\
+24219022671055626321111109370544217506941658960408\
+07198403850962455444362981230987879927244284909188\
+84580156166097919133875499200524063689912560717606\
+05886116467109405077541002256983155200055935729725\
+71636269561882670428252483600823257530420752963450"
+
+int product(char *ptr, int n) {
+	
+	int i, r = 1;
+	
+	for(i=0; i < n; i++)
+		r *= ((int)ptr[i]) - 0x30;
+	
+	return r;
+}
+
+int main() {
+	
+	int i;
+	char str[] = BIGWALLOFTEXT;
+	int tmp, r = 0;
+	
+	for(i=0; i < N_CHARS-5; i++) {
+		
+		tmp = product(&str[i], 5);
+	
+		if (tmp > r)
+			r = tmp;
+	}
+	
+	printf("%i\n", r);
+	
+	return 0;
+}

p9.c

@@ -0,0 +1,43 @@
+
+/*
+ * http://projecteuler.net
+ * 
+ * Projecteuler - Problem 9
+ * ------------------------
+ * 2010-04-06 Henrik Hautakoski
+ * 
+ * the math stuff here (simple algebra)
+ *
+ *         1000 = a + b + c
+ *         10^6 = (a + b + c)^2 
+ *         10^6 = a^2 + ab + ac + ba + b^2 + bc + ca  + cb + c^2
+ *         10^6 = a^2 + b^2 + c^2 + 2(ab + ac + bc)                    
+ *         10^6 = 2a^2 + 2b^2 + 2(ab + ac + bc)                    (substitution: a^2 + b^2 = c^2, pythagorean theorem)
+ *       10^6/2 = a^2 + b^2 + ab + ac + bc
+ *  10^6/2 + ab = a^2 + b^2 + 2ab + ac + bc
+ *  10^6/2 + ab = (a + b)(a + b) + (a + b)c
+ *  10^6/2 + ab = (a + b + c)(a + b)
+ *  10^6/2 + ab = 1000a + 1000b                                    (substitution: a + b + c = 1000)
+ *       10^6/2 = 1000a + 1000b - ab
+ *       10^6/2 = a(1000 - b) + 1000b
+ *
+ *  Equation to use: a = (10^6/2 - 1000b)/(1000 - b)
+ */
+
+#include <stdio.h>
+
+int main() {
+
+    int a, b;
+    double tmp;
+
+    for(b=1; b < 1000; b++) {
+        a = tmp = (5e5 - 1000*b)/(1000 - b);
+        if (tmp - a >= 0.0 && (tmp - a) < .00000001) {
+            printf("%i\n", a*b*(1000 - (a+b)));
+            break;
+        }
+    }
+
+    return 0;
+}