瞎扯淡 计算某天是一年的第多少天
Python Programming: An Introduction to Computer Science 第 7 章第 15 题：
The days of the year are oftennumberedfrom 1 to through365(or 366). This number can be computed in three steps using int arithmetic: (a) dayNum=31*(month-1)+day (b) if the month is after February, subtract (4*month+23)/10 from dayNum (c) if it’s a leap year and after February 29, add 1
这个方法好巧妙，为什么第 2 步要那样算
第 1 步把每个月都当成 31 天，算多了，所以第 2 步要减少，减少多少呢？很可惜是不规则的...
我们找个最简单的函数，利用整数算术压缩表示这个减少的天数，就假设是 (a*month + b)/c 吧
@months = *3..12
acc = 0
# 2 3 4 5 6 7 8 9 10 11 12
@days_diffs = [3, 0, 1, 0, 1, 0, 0, 1, 0, 1].map do |e|
acc += e
end
def valid? a, b, c
@months.zip @days_diffs do |x, y|
return false if (a * x + b)/c != y
end
true
end
valid_nums = []
0.upto 100 do |a|
0.upto 100 do |b|
1.upto 100 do |c|
if valid?(a, b, c)
valid_nums << [a, b, c]
end
end
end
end
p \
valid_nums.sort_by {|(a, b, c)| a + b + c }
可以看到，其实第 2 步的公式不是最巧妙的，排名第 6 ...
[[2, 11, 5], [3, 14, 7], [4, 17, 9], [4, 22, 10], [5, 20, 11], [4, 23, 10], ...
第 2 步这么写更简单
(b) if the month is after February, subtract (2*month+11)/5 from dayNum