reformat to adhere to PEP8 style

A Chocolate Fiesta.py

@@ -1,75 +1,75 @@
-# -*- coding: utf-8 -*-
-"""
-Problem Statement
-
-Welcome to the exciting class of Professor Manasa. In each lecture she used to play some game while teaching a new
-concept. Today's topic is Set Theory. For today's game, she had given a set A = {a1, a2, ...aN} of N integers to her
-students and asked them to play the game as follows.
-
-At each step of the game she calls a random student and asks him/her to select a non-empty subset from set A such that
-this subset had not been selected earlier and the sum of subset should be even. This game ends when all possible subsets
-had been selected. Manasa needs your help in counting the total number of times students can be called assuming each
-student gives the right answer. While it is given that if two numbers are same in the given set, they have different
-colors. It means that if a1 = a2, then choosing a1 and choosing a2 will be considered as different sets.
-"""
-__author__ = 'Danyang'
-MOD = 10**9+7
-
-
-class Solution(object):
-    def solve_error(self, cipher):
-        """
-        count odd number and even number
-        :param cipher: the cipher
-        """
-        N, lst = cipher
-        odd_cnt = len(filter(lambda x: x%2==1, lst))
-        even_cnt = N-odd_cnt
-
-        a = (2**even_cnt)%MOD
-
-        result = 0
-        result += a-1
-
-        i = 2
-        comb = (odd_cnt)*(odd_cnt-1)/(1*2)
-        while i<=odd_cnt:
-            result += comb*a
-            result %= MOD
-            i += 2
-            comb *= (odd_cnt-i+1)*(odd_cnt-i)/((i-1)*i)
-            comb %= MOD
-
-        return result
-
-    def solve(self, cipher):
-        """
-        count odd number and even number
-        :param cipher: the cipher
-        """
-        N, lst = cipher
-        odd_cnt = len(filter(lambda x: x%2==1, lst))
-        even_cnt = N-odd_cnt
-
-        a = (2**even_cnt)%MOD
-        b = (2**(odd_cnt-1))%MOD
-
-        if odd_cnt!=0:
-            result = a-1+(b-1)*a
-        else:
-            result = a-1
-
-        return result%MOD
-
-
-if __name__=="__main__":
-    import sys
-
-    f = open("1.in", "r")
-    # f = sys.stdin
-    N = int(f.readline().strip())
-    lst = map(int, f.readline().strip().split(' '))
-    cipher = N, lst
-    # solve
-    s = "%s\n"%(Solution().solve(cipher))
-    print s,
+# -*- coding: utf-8 -*-
+"""
+Problem Statement
+
+Welcome to the exciting class of Professor Manasa. In each lecture she used to play some game while teaching a new
+concept. Today's topic is Set Theory. For today's game, she had given a set A = {a1, a2, ...aN} of N integers to her
+students and asked them to play the game as follows.
+
+At each step of the game she calls a random student and asks him/her to select a non-empty subset from set A such that
+this subset had not been selected earlier and the sum of subset should be even. This game ends when all possible subsets
+had been selected. Manasa needs your help in counting the total number of times students can be called assuming each
+student gives the right answer. While it is given that if two numbers are same in the given set, they have different
+colors. It means that if a1 = a2, then choosing a1 and choosing a2 will be considered as different sets.
+"""
+__author__ = 'Danyang'
+MOD = 10 ** 9 + 7
+
+
+class Solution(object):
+    def solve_error(self, cipher):
+        """
+        count odd number and even number
+        :param cipher: the cipher
+        """
+        N, lst = cipher
+        odd_cnt = len(filter(lambda x: x % 2 == 1, lst))
+        even_cnt = N - odd_cnt
+
+        a = (2 ** even_cnt) % MOD
+
+        result = 0
+        result += a - 1
+
+        i = 2
+        comb = (odd_cnt) * (odd_cnt - 1) / (1 * 2)
+        while i <= odd_cnt:
+            result += comb * a
+            result %= MOD
+            i += 2
+            comb *= (odd_cnt - i + 1) * (odd_cnt - i) / ((i - 1) * i)
+            comb %= MOD
+
+        return result
+
+    def solve(self, cipher):
+        """
+        count odd number and even number
+        :param cipher: the cipher
+        """
+        N, lst = cipher
+        odd_cnt = len(filter(lambda x: x % 2 == 1, lst))
+        even_cnt = N - odd_cnt
+
+        a = (2 ** even_cnt) % MOD
+        b = (2 ** (odd_cnt - 1)) % MOD
+
+        if odd_cnt != 0:
+            result = a - 1 + (b - 1) * a
+        else:
+            result = a - 1
+
+        return result % MOD
+
+
+if __name__ == "__main__":
+    import sys
+
+    f = open("1.in", "r")
+    # f = sys.stdin
+    N = int(f.readline().strip())
+    lst = map(int, f.readline().strip().split(' '))
+    cipher = N, lst
+    # solve
+    s = "%s\n" % (Solution().solve(cipher))
+    print s,

A journey to the Moon.py

@@ -1,80 +1,80 @@
-"""
-Problem Statement
-
-The member states of the UN are planning to send 2 people to the Moon. But there is a problem. In line with their
-principles of global unity, they want to pair astronauts of 2 different countries.
-
-There are N trained astronauts numbered from 0 to N-1. But those in charge of the mission did not receive information
-about the citizenship of each astronaut. The only information they have is that some particular pairs of astronauts
-belong to the same country.
-"""
-__author__ = 'Danyang'
-
-
-class DisjointSet(object):
-    def __init__(self, n):
-        self.rank = [1 for _ in xrange(n)]
-        self.parent = [i for i in xrange(n)]
-        self.n = n
-
-    def find(self, i):
-        if i==self.parent[i]:
-            return i
-        else:
-            self.parent[i] = self.find(self.parent[i])  # directly point to ancestor
-            return self.parent[i]
-
-    def union(self, i, j):
-        x = self.find(i)
-        y = self.find(j)
-
-        if x==y: return
-        self.parent[x] = y
-        self.rank[y] += self.rank[x]
-
-    def card(self):
-        card = 0
-        for i in xrange(self.n):
-            if self.parent[i]==i:
-                card += 1
-        return card
-
-
-class Solution(object):
-    def solve(self, cipher):
-        """
-        Disjoint-set data structure
-
-        Union-Find Algorithm
-        :param cipher: the cipher
-        """
-        N, pairs = cipher
-        djs = DisjointSet(N)
-        for a, b in pairs:
-            djs.union(a, b)
-
-        result = 0
-        for i in xrange(N):
-            if djs.find(i)==i:  # set representative
-                # result += (djs.rank[i])*(N-djs.rank[i])/2  # otherwise rounding error
-                result += (djs.rank[i])*(N-djs.rank[i])
-        return result/2
-
-
-if __name__=="__main__":
-    import sys
-
-    f = open("1.in", "r")
-    # f = sys.stdin
-    solution = Solution()
-
-    N, I = map(int, f.readline().strip().split(' '))
-
-    pairs = []
-    for i in xrange(I):
-        pairs.append(map(int, f.readline().strip().split(' ')))
-
-    cipher = N, pairs
-    # solve
-    s = "%s\n"%(solution.solve(cipher))
-    print s,
+"""
+Problem Statement
+
+The member states of the UN are planning to send 2 people to the Moon. But there is a problem. In line with their
+principles of global unity, they want to pair astronauts of 2 different countries.
+
+There are N trained astronauts numbered from 0 to N-1. But those in charge of the mission did not receive information
+about the citizenship of each astronaut. The only information they have is that some particular pairs of astronauts
+belong to the same country.
+"""
+__author__ = 'Danyang'
+
+
+class DisjointSet(object):
+    def __init__(self, n):
+        self.rank = [1 for _ in xrange(n)]
+        self.parent = [i for i in xrange(n)]
+        self.n = n
+
+    def find(self, i):
+        if i == self.parent[i]:
+            return i
+        else:
+            self.parent[i] = self.find(self.parent[i])  # directly point to ancestor
+            return self.parent[i]
+
+    def union(self, i, j):
+        x = self.find(i)
+        y = self.find(j)
+
+        if x == y: return
+        self.parent[x] = y
+        self.rank[y] += self.rank[x]
+
+    def card(self):
+        card = 0
+        for i in xrange(self.n):
+            if self.parent[i] == i:
+                card += 1
+        return card
+
+
+class Solution(object):
+    def solve(self, cipher):
+        """
+        Disjoint-set data structure
+
+        Union-Find Algorithm
+        :param cipher: the cipher
+        """
+        N, pairs = cipher
+        djs = DisjointSet(N)
+        for a, b in pairs:
+            djs.union(a, b)
+
+        result = 0
+        for i in xrange(N):
+            if djs.find(i) == i:  # set representative
+                # result += (djs.rank[i])*(N-djs.rank[i])/2  # otherwise rounding error
+                result += (djs.rank[i]) * (N - djs.rank[i])
+        return result / 2
+
+
+if __name__ == "__main__":
+    import sys
+
+    f = open("1.in", "r")
+    # f = sys.stdin
+    solution = Solution()
+
+    N, I = map(int, f.readline().strip().split(' '))
+
+    pairs = []
+    for i in xrange(I):
+        pairs.append(map(int, f.readline().strip().split(' ')))
+
+    cipher = N, pairs
+    # solve
+    s = "%s\n" % (solution.solve(cipher))
+    print s,