Project Euler Solution

project_euler/Problem 04/sol.c

@@ -0,0 +1,31 @@
+#include <stdio.h>
+
+int is_palindromic(unsigned int n)
+{
+    unsigned int reversed = 0, t = n;
+
+    while (t>0)
+    {
+        reversed = 10 * reversed + (t % 10);
+        t /= 10;
+    }
+    return reversed == n;
+}
+
+int main(void)
+{
+    unsigned int i, j, max = 0;
+    for (i = 100; i <= 999; i++)
+    {
+        for (j = 100; j <= 999; j++)
+        {
+            unsigned int p = i * j;
+            if (is_palindromic(p) && p > max)
+            {
+                max = p;
+            }
+        }
+    }
+    printf("%u\n", max);
+    return 0;
+}

project_euler/Problem 05/sol.c

@@ -0,0 +1,30 @@
+#include <stdio.h>
+
+unsigned long gcd(unsigned long a, unsigned long b) {
+    unsigned long r;
+    if (a > b) {
+        unsigned long t = a;
+        a = b;
+        b = t;
+    }
+    while (r = a % b) {
+        a = b;
+        b = r;
+    }
+    return b;
+}
+
+unsigned long lcm(unsigned long a, unsigned long b) {
+    unsigned long long p = (unsigned long long)a * b;
+    return p / gcd(a, b);
+}
+
+int main(void) {
+    unsigned long ans = 1;
+    unsigned long i;
+    for (i = 1; i <= 20; i++) {
+        ans = lcm(ans, i);
+    }
+    printf("%lu\n", ans);
+    return 0;
+}

project_euler/Problem 06/sol.c

@@ -0,0 +1,12 @@
+#include <stdio.h>
+
+int main(void) {
+    unsigned s1 = 0, s2 = 0, i;
+    for (i = 1; i <= 100; i++) {
+        s1 += i * i;
+        s2 += i;
+    }
+    unsigned ans = s2 * s2 - s1;
+    printf("%u\n", ans);
+    return 0;
+}

project_euler/Problem 07/sol.c

@@ -0,0 +1,27 @@
+#include <stdio.h>
+#include <stdlib.h>
+
+int main(void) {
+    char *sieve;
+    size_t i;
+    unsigned count = 0;
+    size_t n = 1000000;
+    const unsigned target = 10001;
+
+    sieve = calloc(n, sizeof *sieve);
+    for (i = 2; i < n; i++) {
+        if (!sieve[i]) {
+            size_t j; 
+            count++;
+            if (count == target) {
+                printf("%lu\n", i);
+                break;
+            }
+            for (j = i * 2; j < n; j += i) {
+                sieve[j] = 1;
+            }
+        }
+    }
+    free(sieve);
+    return 0;
+}