add problem2[0-9]

20-29/problem20.c

@@ -0,0 +1,30 @@
+/*
+ * Copyright (c) 2009, eagletmt
+ * Released under the MIT License <http://opensource.org/licenses/mit-license.php>
+ */
+
+#include <stdio.h>
+#include <stdlib.h>
+#include <gmp.h>
+
+int main(void)
+{
+  mpz_t n;
+  char *str;
+  int i;
+  int sum = 0;
+
+  mpz_init(n);
+  mpz_fac_ui(n, 100);
+  str = mpz_get_str(NULL, 10, n);
+  for (i = 0; str[i]; i++) {
+    sum += str[i]-'0';
+  }
+  printf("%d\n", sum);
+
+  free(str);
+  mpz_clear(n);
+
+  return 0;
+}
+

20-29/problem21.c

@@ -0,0 +1,40 @@
+/*
+ * Copyright (c) 2009, eagletmt
+ * Released under the MIT License <http://opensource.org/licenses/mit-license.php>
+ */
+
+#include <stdio.h>
+
+static int divisors_sum(int n);
+
+int main(void)
+{
+  int i, d, sum = 0;
+
+  for (i = 2; i < 10000; i++) {
+    d = divisors_sum(i);
+    if (i < d && i == divisors_sum(d)) {
+      sum += i + d;
+    }
+  }
+  printf("%d\n", sum);
+
+  return 0;
+}
+
+int divisors_sum(int n)
+{
+  int sum= 1;
+  int i, k = n;
+
+  for (i = 2; i <= k; i++) {
+    int p = 1;
+    while (k % i == 0) {
+      p *= i;
+      k /= i;
+    }
+    sum *= (p*i - 1)/(i-1);
+  }
+  return sum - n;
+}
+

20-29/problem22.c

@@ -0,0 +1,59 @@
+/*
+ * Copyright (c) 2009, eagletmt
+ * Released under the MIT License <http://opensource.org/licenses/mit-license.php>
+ */
+
+#include <stdio.h>
+#include <stdlib.h>
+#include <string.h>
+#include <assert.h>
+
+static int compare(const void *x, const void *y);
+
+int main(void)
+{
+  int count = 0;
+  FILE *fp;
+  char path[] = "names.txt";
+  char **names;
+  int i;
+  int sum = 0;
+
+  fp = fopen(path, "r");
+  if (!fp) {
+    perror(path);
+    return 1;
+  }
+
+  /* parse names.txt */
+  names = malloc(8192 * sizeof *names);
+  names[count] = malloc(16);
+  while (fscanf(fp, "\"%[^\"]\",", names[count]) != EOF) {
+    names[++count] = malloc(16);
+  }
+  free(names[count]);
+  fclose(fp);
+
+  qsort(names, count, sizeof *names, compare);
+
+  for (i = 0; i < count; i++) {
+    int j, s = 0;
+    for (j = 0; names[i][j] != 0; j++) {
+      s += names[i][j]-'A'+1;
+    }
+    sum += s * (i+1);
+
+    free(names[i]);
+  }
+  free(names);
+
+  printf("%d\n", sum);
+
+  return 0;
+}
+
+int compare(const void *x, const void *y)
+{
+  return strcmp(*(char * const *)x, *(char * const *)y);
+}
+

20-29/problem23.c

@@ -0,0 +1,71 @@
+/*
+ * Copyright (c) 2009, eagletmt
+ * Released under the MIT License <http://opensource.org/licenses/mit-license.php>
+ */
+
+#include <stdio.h>
+#include <stdlib.h>
+
+static int is_abundant(int n);
+
+int main(void)
+{
+  int i, j;
+  const int upper_limit = 28123;
+  int *abundants, abundants_count = 0;
+  char *expressed;
+  int sum = 0;
+
+  for (i = 1; i < upper_limit; i++) {
+    if (is_abundant(i)) {
+      abundants_count++;
+    }
+  }
+
+  abundants = malloc(abundants_count * sizeof *abundants);
+  j = 0;
+  for (i = 1; i < upper_limit; i++) {
+    if (is_abundant(i)) {
+      abundants[j++] = i;
+    }
+  }
+
+  expressed = calloc(upper_limit, sizeof *expressed);
+  for (i = 0; i < abundants_count; i++) {
+    for (j = i; j < abundants_count; j++) {
+      int k = abundants[i] + abundants[j];
+      if (k >= upper_limit) {
+        break;
+      }
+      expressed[k-1] = 1;
+    }
+  }
+
+  for (i = 1; i < upper_limit; i++) {
+    if (!expressed[i-1]) {
+      sum += i;
+    }
+  }
+  printf("%d\n", sum);
+
+  free(abundants);
+  free(expressed);
+
+  return 0;
+}
+
+int is_abundant(int n)
+{
+  int sum = 1;
+  int i, k = n;
+  for (i = 2; i <= k; i++) {
+    int p = 1;
+    while (k % i == 0) {
+      p *= i;
+      k /= i;
+    } 
+    sum *= (p*i - 1)/(i-1);
+  }
+  return 2*n < sum;
+}
+

20-29/problem24.cc

@@ -0,0 +1,28 @@
+/*
+ * Copyright (c) 2009, eagletmt
+ * Released under the MIT License <http://opensource.org/licenses/mit-license.php>
+ */
+
+#include <iostream>
+#include <algorithm>
+
+#define N 1000000
+
+int main(void)
+{
+  using namespace std;
+
+  int a[] = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9};
+  int i;
+
+  for (i = 0; i <= N; i++) {
+    next_permutation(a, a + sizeof a/sizeof *a);
+  }
+  for (i = 0; i < 10; i++) {
+    cout << a[i];
+  }
+  cout << endl;
+
+  return 0;
+}
+

20-29/problem25.c

@@ -0,0 +1,36 @@
+/*
+ * Copyright (c) 2009, eagletmt
+ * Released under the MIT License <http://opensource.org/licenses/mit-license.php>
+ */
+
+#include <stdio.h>
+#include <stdlib.h>
+#include <string.h>
+#include <gmp.h>
+
+int main(void)
+{
+  mpz_t n;
+  int i = 1;
+
+  mpz_init(n);
+
+  for (i = 1;; i++) {
+    char *str;
+    size_t len;
+
+    mpz_fib_ui(n, i);
+    str = mpz_get_str(NULL, 10, n);
+    len = strlen(str);
+    free(str);
+    if (len == 1000) {
+      printf("%d\n", i);
+      break;
+    }
+  }
+
+  mpz_clear(n);
+
+  return 0;
+}
+

20-29/problem26.c

@@ -0,0 +1,43 @@
+/*
+ * Copyright (c) 2009, eagletmt
+ * Released under the MIT License <http://opensource.org/licenses/mit-license.php>
+ */
+
+#include <stdio.h>
+#include <gmp.h>
+
+#define M 1000
+
+int main(void)
+{
+  mpz_t p, r;
+  unsigned maxlen = 0, imax = 0;
+  unsigned i;
+
+  mpz_init(p); mpz_init(r);
+
+  for (i = 2; i < M; i++) {
+    unsigned len = 1;
+
+    if (i % 2 == 0 || i % 5 == 0) {
+      continue;
+    }
+
+    mpz_set_ui(p, 10);
+    while (mpz_mod_ui(r, p, i), mpz_cmp_ui(r, 1) != 0) {
+      len++;
+      mpz_mul_ui(p, p, 10);
+    }
+    if (len > maxlen) {
+      maxlen = len;
+      imax = i;
+    }
+  }
+  printf("%u\n", imax);
+
+  mpz_clear(p);
+  mpz_clear(r);
+
+  return 0;
+}
+

20-29/problem27.c

@@ -0,0 +1,49 @@
+/*
+ * Copyright (c) 2009, eagletmt
+ * Released under the MIT License <http://opensource.org/licenses/mit-license.php>
+ */
+
+#include <stdio.h>
+#include <math.h>
+
+static int is_prime(int n);
+
+int main(void)
+{
+  int a, b, n;
+  int nmax = 0, amax = 0, bmax = 0;
+
+  for (a = -999; a < 1000; a++) {
+    for (b = -999; b < 1000; b++) {
+      for (n = 0; is_prime(n*n + a*n + b); n++);
+      if (n > nmax) {
+        nmax = n;
+        amax = a;
+        bmax = b;
+      }
+    }
+  }
+  printf("%d\n", amax * bmax);
+
+  return 0;
+}
+
+int is_prime(int n)
+{
+  int i;
+
+  if (n <= 1) {
+    return 0;
+  }
+  if (n == 2) {
+    return 1;
+  }
+
+  for (i = 2; i < sqrt((double)n); i++) {
+    if (n % i == 0) {
+      return 0;
+    }
+  }
+  return 1;
+}
+

20-29/problem28.c

@@ -0,0 +1,19 @@
+/*
+ * Copyright (c) 2009, eagletmt
+ * Released under the MIT License <http://opensource.org/licenses/mit-license.php>
+ */
+
+#include <stdio.h>
+
+#define N 1001
+
+int main(void)
+{
+  int sum = 1, i;
+  for (i = 3; i <= N; i += 2) {
+    sum += 2*(2*i*i - 3*i + 3);
+  }
+  printf("%d\n", sum);
+  return 0;
+}
+