原文:http://lua-users.org/wiki/FunWithTables
题目:输出1到100之间的数哪些能被3整除,哪些能被5整除.
常规的做法是做一个从1到100的循环,让数模3和5,使用if else if这样语句进行判断.
for i = 1, 100 do
if i %3 == 0 and i % 5 == 0 then
print(i .. ' is divisible by both 3 and 5!')
elseif i % 3 ~= 0 and i % 5 ~= 0 then
print(i .. ' is not divisible by either 3 or 5!')
elseif i % 3 == 0 then
print(i .. ' is divisible by 3 only!')
elseif i % 5 == 0 then
print(i .. ' is divisible by 5 only!')
end
end
一个有趣的做法是使用表.因为一个数要么能被3整除要么不被3整除,要么被5整除要么不能整除.
那么一个数能否被3和5整除一共是4种情况.分别用true和false来表示能整除和不能整除,那么结果是:
能被3整除 对 能被5整除/不能被5整除
true 对 true/false)
能被5整除 对 能被3整除/不能被3整除
false 对 true/false
这相当于一个二维数组,第一维是两个元素:true或false ,第二维也是两个元素:true或false.
Lua处理数组是非常强大的,把他们放到表里就是:
{ ['true']={'true','false'} }
那么上述题目便有了如下的写法:
result={
[true] = { [true] = ' is divisible by both 3 and 5!', [false ] = ' is divisible by 3 only!' },
[false] = { [true] = ' is divisible by 5 only!', [false] = ' is not divisible by either 3 or 5!' }
}
for i=1,100 do
print(i .. result[i % 3 == 0][i % 5 == 0] )
end
其他演化的方法有:
-- common to both variants
local a, b, c, d =
" is not divisible by either 3 or 5",
" is divisible by 3 but not by 5",
" is divisible by 5 but not by 3",
" is divisible by both 3 and 5"
-- variant #1 --
local t = { a,b,c,d }
for x=1,100 do
print(x .. t[(x%3==0 and 1 or 0) + (x%5==0 and 2 or 0) + 1])
end
-- variant #2 --
for x=1,100 do
print(x .. (x%3==0 and (x%5==0 and d or b) or (x%5==0 and c or a)))
end
文档信息
- 本文作者：zhupite
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