Added Problem 18

Problem_018.py

@@ -0,0 +1,79 @@
+######################################################################
+# By starting at the top of the triangle below and moving to adjacent numbers on the
+# row below, the maximum total from top to bottom is 23.
+#
+# 3
+# 7 4
+# 2 4 6
+# 8 5 9 3
+#
+# That is, 3 + 7 + 4 + 9 = 23.
+#
+# Find the maximum total from top to bottom of the triangle below:
+#
+# 75
+# 95 64
+# 17 47 82
+# 18 35 87 10
+# 20 04 82 47 65
+# 19 01 23 75 03 34
+# 88 02 77 73 07 63 67
+# 99 65 04 28 06 16 70 92
+# 41 41 26 56 83 40 80 70 33
+# 41 48 72 33 47 32 37 16 94 29
+# 53 71 44 65 25 43 91 52 97 51 14
+# 70 11 33 28 77 73 17 78 39 68 17 57
+# 91 71 52 38 17 14 91 43 58 50 27 29 48
+# 63 66 04 68 89 53 67 30 73 16 69 87 40 31
+# 04 62 98 27 23 09 70 98 73 93 38 53 60 04 23
+#
+# NOTE: As there are only 16384 routes, it is possible to solve this problem by trying every route.
+# However, Problem 67, is the same challenge with a triangle containing one-hundred rows; it cannot be solved
+# by brute force, and requires a clever method! ;o)
+######################################################################
+
+triangle = "75 \n\
+95 64 \n\
+17 47 82 \n\
+18 35 87 10 \n\
+20 04 82 47 65 \n\
+19 01 23 75 03 34 \n\
+88 02 77 73 07 63 67 \n\
+99 65 04 28 06 16 70 92 \n\
+41 41 26 56 83 40 80 70 33 \n\
+41 48 72 33 47 32 37 16 94 29 \n\
+53 71 44 65 25 43 91 52 97 51 14 \n\
+70 11 33 28 77 73 17 78 39 68 17 57 \n\
+91 71 52 38 17 14 91 43 58 50 27 29 48 \n\
+63 66 04 68 89 53 67 30 73 16 69 87 40 31 \n\
+04 62 98 27 23 09 70 98 73 93 38 53 60 04 23 \n"
+
+other = "3 \n\
+7 4 \n\
+2 4 6 \n\
+8 5 9 3 \n"
+
+tri = []
+temp = []
+tempstr = ''
+
+for c in triangle:
+    if c is " ":
+        temp.append(int(tempstr))
+        tempstr = ""
+    elif c is "\n":
+        tri.append(temp)
+        temp = []
+    else:
+        tempstr += c
+
+tri.reverse()
+
+for row in range(1, len(tri)):
+    for number in range(0, len(tri[row])):
+        tri[row][number] += max(tri[row - 1][number], tri[row - 1][number + 1])
+
+tri.reverse()
+
+print(tri[0][0])
+# Solution: 1074