From 4c69de67681a1bb6e78c9adf1072a5f3b459e4c9 Mon Sep 17 00:00:00 2001
From: Charles <sircharlesaze@gmail.com>
Date: Sun, 29 Sep 2019 21:37:51 +0200
Subject: problem 145 c

---
 ...reversible_numbers_are_there_below_onebillion.c | 52 ++++++++++++++++++++++
 1 file changed, 52 insertions(+)
 create mode 100644 c/145-how_many_reversible_numbers_are_there_below_onebillion.c

(limited to 'c')

diff --git a/c/145-how_many_reversible_numbers_are_there_below_onebillion.c b/c/145-how_many_reversible_numbers_are_there_below_onebillion.c
new file mode 100644
index 0000000..4cddb5e
--- /dev/null
+++ b/c/145-how_many_reversible_numbers_are_there_below_onebillion.c
@@ -0,0 +1,52 @@
+/*
+* How many reversible numbers are there below one-billion?
+* Problem 145
+*
+* Some positive integers n have the property that the sum [ n + reverse(n) ] consists entirely of odd (decimal) digits. For instance, 36 + 63 = 99 and 409 + 904 = 1313. We will call such numbers reversible; so 36, 63, 409, and 904 are reversible. Leading zeroes are not allowed in either n or reverse(n).
+* There are 120 reversible numbers below one-thousand.
+* How many reversible numbers are there below one-billion (109)?
+*/
+
+#include <stdio.h>
+#include <stdbool.h>
+
+int reverse(int n);
+bool is_reversible(int n);
+
+int main(void)
+{
+    int counter = 0;
+    for (int i = 0; i < 1000000000; i++)
+    {
+        if (i % 10 == 0)
+            continue;
+        if (is_reversible(i))
+            counter++;
+    }
+    printf("%d\n", counter);
+    return 0;
+}
+
+bool is_reversible(int n)
+{
+    int sum = n + reverse(n);
+    while (sum > 0)
+    {
+        if ((sum % 10) % 2 == 0)
+            return false;
+        sum /= 10;
+    }
+    return true;
+}
+
+int reverse(int n)
+{
+    int rn = 0;
+    while (n > 0)
+    {
+        rn *= 10;
+        rn += n % 10;
+        n /= 10;
+    }
+    return rn;
+}
-- 
cgit