计算质数和回文数的小程序

Ruby 代码如下

def isZhishu?(num)


[code]count = 2

while count < num do

return false if ( num % count ) == 0

count = count + 1

end

return true

end

def isHuiwen?(num)

s = num.to_s.reverse.to_i

return true if num == s

return false

end

count = 0

10000.times{ |i|

i = i + 1

if isZhishu?(i) && isHuiwen?(i) then

printf "%4d", i

print "\n"

count = count + 1

end

}

print "\n\nTotal:#{count}"

[/code]

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上面这个方法非常笨重，时间复杂度是 O（n^2），可以进行一些优化。根据 @sdpfoue 的建议，做了优化。

首先就是可以只对大于3的奇数进行检查，因为偶数肯定可以被2整除，所以不需要考虑。

另外循环相除的时候，可以只除以质数，这样也能够减少不少步骤。但是会增加空间的消耗，就是所谓的用空间换时间。

具体代码如下：

def isZhishu?(num, arrZhishu)


[code]return true if num == 1 || num == 2

count = 2

if( arrZhishu.empty? )then

#count = 2

    while count < num do

return false if ( num % count ) == 0

if( count >= 11 ) then

count = count + 2 # Only judge even numbers

else

count = count + 1

end

end

    return true

else

arrZhishu.each{|value|

next if value == 1

    return false if ( num % value ) == 0

}

    return true

end

end

def isHuiwen?(num)

s = num.to_s.reverse.to_i

return true if num == s

return false

end

count = 0

arrZhishu = Array.new

i = 0

while i < 10000000 do

if i >= 11 then

i = i + 2

else

    i = i + 1

end

if isZhishu?(i, arrZhishu) && isHuiwen?(i) then

arrZhishu.push(i)

#printf "%4d", i

#print "\n"

count = count + 1

end

end

print "\n\nTotal:#{count}"

[/code]

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