Merge pull request TheAlgorithms#268 from daniel-s-ingram/master

Solution to Problem 48

Author: harshildarji

Project Euler/Problem 17/sol1.py

@@ -0,0 +1,35 @@
+from __future__ import print_function
+'''
+Number letter counts
+Problem 17
+
+If the numbers 1 to 5 are written out in words: one, two, three, four, five, then there are 3 + 3 + 5 + 4 + 4 = 19 letters used in total.
+
+If all the numbers from 1 to 1000 (one thousand) inclusive were written out in words, how many letters would be used?
+
+
+NOTE: Do not count spaces or hyphens. For example, 342 (three hundred and forty-two) contains 23 letters and 115 (one hundred and fifteen) 
+contains 20 letters. The use of "and" when writing out numbers is in compliance with British usage.
+'''
+
+ones_counts = [0, 3, 3, 5, 4, 4, 3, 5, 5, 4, 3, 6, 6, 8, 8, 7, 7, 9, 8, 8] #number of letters in zero, one, two, ..., nineteen (0 for zero since it's never said aloud)
+tens_counts = [0, 0, 6, 6, 5, 5, 5, 7, 6, 6] #number of letters in twenty, thirty, ..., ninety (0 for numbers less than 20 due to inconsistency in teens)
+
+count = 0
+
+for i in range(1, 1001):
+	if i < 1000:
+		if i >= 100:
+			count += ones_counts[i/100] + 7 #add number of letters for "n hundred"
+
+			if i%100 is not 0:
+				count += 3 #add number of letters for "and" if number is not multiple of 100
+
+		if 0 < i%100 < 20:
+			count += ones_counts[i%100] #add number of letters for one, two, three, ..., nineteen (could be combined with below if not for inconsistency in teens)
+		else:
+			count += ones_counts[i%10] + tens_counts[(i%100-i%10)/10] #add number of letters for twenty, twenty one, ..., ninety nine
+	else:
+		count += ones_counts[i/1000] + 8
+
+print(count)

Project Euler/Problem 48/sol1.py

@@ -0,0 +1,21 @@
+from __future__ import print_function
+'''
+Self Powers
+Problem 48
+
+The series, 11 + 22 + 33 + ... + 1010 = 10405071317.
+
+Find the last ten digits of the series, 11 + 22 + 33 + ... + 10001000.
+'''
+
+try:
+	xrange
+except NameError:
+	xrange = range
+
+total = 0
+for i in xrange(1, 1001):
+	total += i**i
+
+
+print(str(total)[-10:])