SPOJ Problems

Author: OmkarPathak

CompetitiveProgramming/SPOJ/P03_ACPC10A.py

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+# According to Wikipedia, an arithmetic progression (AP) is a sequence of numbers such that the difference of any two
+# successive members of the sequence is a constant. For instance, the sequence 3, 5, 7, 9, 11, 13, . . . is an arithmetic
+# progression with common difference 2. For this problem, we will limit ourselves to arithmetic progression whose common
+# difference is a non-zero integer.
+# On the other hand, a geometric progression (GP) is a sequence of numbers where each term after the first is found by
+# multiplying the previous one by a fixed non-zero number called the common ratio. For example, the sequence 2, 6, 18, 54, . . .
+# is a geometric progression with common ratio 3. For this problem, we will limit ourselves to geometric progression whose
+# common ratio is a non-zero integer.
+# Given three successive members of a sequence, you need to determine the type of the progression and the next successive
+# member.
+#
+# Input
+#
+# Your program will be tested on one or more test cases. Each case is specified on a single line with three integers
+# (−10, 000 < a1 , a2 , a3 < 10, 000) where a1 , a2 , and a3 are distinct.
+# The last case is followed by a line with three zeros.
+#
+# Output
+#
+# For each test case, you program must print a single line of the form:
+# XX v
+# where XX is either AP or GP depending if the given progression is an Arithmetic or Geometric Progression. v is the next member of the given sequence. All input cases are guaranteed to be either an arithmetic or geometric progressions.
+#
+# Example
+#
+# Input:
+# 4 7 10
+# 2 6 18
+# 0 0 0
+#
+# Output:
+# AP 13
+# GP 54
+
+while(True):
+    a, b, c = map(int, input().split())
+    if a == b == c == 0:
+        break
+    if b-a == c-b:
+        print ('AP', c + b - a)
+    else:
+        print ('GP', c * (b // a))