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diff --git a/python/026-reciprocal_cycles.py b/python/026-reciprocal_cycles.py
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+# ###
+# Reciprocal cycles
+# Problem 26
+# 
+# A unit fraction contains 1 in the numerator. The decimal representation of the unit fractions with denominators 2 to 10 are given:
+# 
+# 1/2= 0.5
+# 1/3= 0.(3)
+# 1/4= 0.25
+# 1/5= 0.2
+# 1/6= 0.1(6)
+# 1/7= 0.(142857)
+# 1/8= 0.125
+# 1/9= 0.(1)
+# 1/10= 0.1
+# 
+# Where 0.1(6) means 0.166666..., and has a 1-digit recurring cycle. It can be seen that 1/7 has a 6-digit recurring cycle.
+# Find the value of d < 1000 for which 1/d contains the longest recurring cycle in its decimal fraction part.
+# ###
+
+
+import decimal
+
+
+MIN_CYCLE = 3
+
+def get_start_cycle(s):
+    cycle = s[0]
+    detected_counter = 0
+    i = 1
+    while i < len(s):
+        if detected_counter >= MIN_CYCLE:
+            return cycle
+        if s[i:].find(cycle) == 0:
+            detected_counter += 1
+            i += len(cycle) - 1
+        else:
+            detected_counter = 0
+            cycle += s[i]
+        i += 1
+    return ""
+
+def get_cycle(s):
+    for j in range(len(s)):
+        cycle = get_start_cycle(s[j:])
+        if cycle != "":
+            return cycle
+    return ""
+
+   
+if __name__ == "__main__":
+    decimal.getcontext().prec = 4000
+    dec_one = decimal.Decimal(1)
+    unit_fracs = [str(dec_one / decimal.Decimal(x))[2:] for x in range(2, 1000)]
+    cycles = [(n, get_cycle(x)) for n, x in zip(range(2, 1000), unit_fracs)]
+    print(sorted(cycles, key=lambda t: len(t[1]), reverse=True)[0][0])