Review of Codeforces 6B and 6C-优快云博客

本文链接：https://blog.youkuaiyun.com/superturbofsi/article/details/41171141

本文介绍使用Python解决两个问题的方法：一是利用集合(set)属性找出总统办公室的颜色方案；二是通过贪心算法分配巧克力以确保Alice和Bob之间的公平竞争。文章提供了源代码及详细解释。

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This question is about using 'set' in pyhon. set has a property that it only add elements that not included in the current set. which is an ideal tool in this question.
First add '.' and 'C(the president's color)' in the set and then search the matrix and when it find 'C', it search around the element and add the alphabeta that not included in the set.

The source code is:

n, m, c = raw_input().split()
n, m = map(int, [n, m])
a = [raw_input() for i in xrange(n)]

s = set(['.', c])
for i in xrange(n):
	for j in xrange(m):
		if a[i][j] == c:
			for x, y in [(i - 1, j), (i + 1, j), (i, j - 1), (i, j + 1)]:
				if x >= 0 and x < n and y >= 0 and y < m: s.add(a[x][y])

print len(s) - 2

I also have an other idea, but is trival and has wrong answer. I dont't know why.

6C. Alice, Bob and Chocolate

This question is about greedy algorithm. The only thing you need to remind is there are corner cases in this task when n== 1 and n == 2. every time compare the time consumed by alice and bob, and add one chocolate to the fewer one. The source code as follows:

n = input()
a = list(map(int, raw_input().split()))

i = 0
j = n-1
m = a[i]
k = a[j]


if n == 1:
	print 1, 0
elif n == 2:
	print 1, 1
else:
	while j-i != 1:
		if (m <= k):
			i += 1
			m += a[i]
		elif (m > k):
			j -= 1
			k += a[j]
	print i+1, n - j

also, you need to be care the output is i+1 and n-j