blocks|key|5255438|text|xrange()的第三个参数是增量值。默认值为1，这意味着计数器将以递增的方向计数。要以递减方向计数，请使用-1。您的a计数器将从999变为101+(+xrange()迭代器在到达第二个参数值之前停止)。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|5255439|有关将来的参考，请参阅xrange()+documentation。|5255440|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.python.org/library/functions.html#xrange^0|0|8|1N|1|24|8|0|B|8|B|M|0|0^^$0|@$1|2|3|4|5|6|7|P|8|@$9|Q|A|R|B|C]|$9|S|A|T|B|C]|$9|U|A|V|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|W|8|@$9|X|A|Y|B|C]]|D|@$9|Z|A|10|1|11]]|E|$]]|$1|H|3|-4|5|6|7|12|8|@]|D|@]|E|$]]]|I|$J|$5|K|L|M|E|$N|O]]]]

The third parameter to <code>xrange()</code> is the increment value. The default is 1, which means the counter will count in an increasing direction. To count in a decreasing direction, use -1. Your <code>a</code> counter will go from 999 to 101 (the <code>xrange()</code> iterator stops just before it reaches the second parameter value).

For future reference, see the <a href="http://docs.python.org/library/functions.html#xrange" rel="nofollow"><code>xrange()</code> documentation</a>.

blocks|key|3143220|text|-1指定了一个负步骤。因此从999下降到100+(独占)。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3143221|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

The -1 specifies a negative step. Thus moving from 999 descending to 100 (exclusive).

blocks|key|5255515|text|xrange函数有三个参数：start、stop和step。它返回一个从start到stop的数字范围，但不包括它。如果'start‘大于stop，则必须提供负步长。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|5255516|所以基本上xrange(999,+100,+-1)会给你[999,+998,+...,+101]|5255517|entityMap^0|0|6|E|5|K|4|P|4|10|5|16|4|1X|4|0|5|K|S|K|0^^$0|@$1|2|3|4|5|6|7|J|8|@$9|K|A|L|B|C]|$9|M|A|N|B|C]|$9|O|A|P|B|C]|$9|Q|A|R|B|C]|$9|S|A|T|B|C]|$9|U|A|V|B|C]|$9|W|A|X|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|Y|8|@$9|Z|A|10|B|C]|$9|11|A|12|B|C]]|D|@]|E|$]]|$1|H|3|-4|5|6|7|13|8|@]|D|@]|E|$]]]|I|$]]

<code>xrange</code> function takes three arguments: <code>start</code>, <code>stop</code> and <code>step</code>.
It returns a range of numbers starting from <code>start</code> continuing to <code>stop</code>, but not including it. If 'start' is bigger than <code>stop</code>, negative step must be provided.

So basically <code>xrange(999, 100, -1)</code> will give you <code>[999, 998, ..., 101]</code>

blocks|key|5255541|text|这意味着您正在以1为增量递减(因此是负号)，从999到100|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5255542|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

It means that you are decreasing (hence the negative sign) in increments of 1, from 999 to 100

blocks|key|279748|text|这是我在项目Euler+#4中得到的问题：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|279749|def+Palindrome(s):
++++if+s+==+s[::-1]:
++++++++return+True
++++else:
++++++++return+False

i+=+100
j+=+100
greatest+=+0
while+(i+<=+999):
++++while+(j+<=+999):
++++++++product+=+i+*+j
++++++++if+(product+>+greatest+and+Palindrome(str(product))):
++++++++++++greatest+=+product
++++++++j+%2B=+1
++++j+=+100
++++i+%2B=+1
print+"Answer:+"+%2B+str(greatest)|code-block|syntax|javascript|279750|-M1K3|279751|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

This is what I got for the project Euler #4 question: 

<pre><code>def Palindrome(s):
 if s == s[::-1]:
 return True
 else:
 return False

i = 100
j = 100
greatest = 0
while (i &lt;= 999):
 while (j &lt;= 999):
 product = i * j
 if (product &gt; greatest and Palindrome(str(product))):
 greatest = product
 j += 1
 j = 100
 i += 1
print "Answer: " + str(greatest)
</code></pre>

-M1K3

blocks|key|3143409|text|我的Python解决方案：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3143410|container+=+[]

for+i+in+range(100,+999):
++++for+j+in+range(100,+999):
++++++++num+=+i+*+j
++++++++if+str(num)+==+str(num)[::-1]:
++++++++++++container.append(num)

print(max(container))|code-block|syntax|javascript|3143411|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

My Python solution:

<pre><code>container = []

for i in range(100, 999):
 for j in range(100, 999):
 num = i * j
 if str(num) == str(num)[::-1]:
 container.append(num)

print(max(container))
</code></pre>

blocks|key|278405|text|__author__+=+'shreysatapara'
f=0
for+a+in+range(100,1000):
++++for+b+in+range(100,1000):
++++++++c=a*b
++++++++d=c
++++++++s=0
++++++++x=0
++++++++for+i+in+range(0,len(str(c))):
++++++++++++s=c%2510
++++++++++++x=x*10%2Bs
++++++++++++c=c//10
++++++++++++if(x==d):
++++++++++++++++if(x>f):
++++++++++++++++++++f=x
++++++++++++++++++++print(f)|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|278406|bingo+answer是你在编译器中的最后一个数字...|unstyled|278407|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|J|8|@]|9|@]|A|$]]|$1|G|3|-4|5|F|7|K|8|@]|9|@]|A|$]]]|H|$]]

<pre><code>__author__ = 'shreysatapara'
f=0
for a in range(100,1000):
 for b in range(100,1000):
 c=a*b
 d=c
 s=0
 x=0
 for i in range(0,len(str(c))):
 s=c%10
 x=x*10+s
 c=c//10
 if(x==d):
 if(x&gt;f):
 f=x
 print(f)
</code></pre>

<h1>bingo answer is your last number in compiler....</h1>

blocks|key|2186593|text|import+time
start_time+=+time.time()
largest_number+=+0
for+i+in+range(999,100,-1):
++++for+j+in+range(i,100,-1):
++++++++string_number+=+str(i+*+j)
++++++++if+string_number+==+string_number[::-1]:
++++++++++++if+i*j+>+largest_number:
++++++++++++++++largest_number+=+i+*+j
print(largest_number)
print(time.time()+-+start_time,"+seconds")|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|2186594|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>import time
start_time = time.time()
largest_number = 0
for i in range(999,100,-1):
 for j in range(i,100,-1):
 string_number = str(i * j)
 if string_number == string_number[::-1]:
 if i*j &gt; largest_number:
 largest_number = i * j
print(largest_number)
print(time.time() - start_time," seconds")
</code></pre>

blocks|key|3143316|text|负的是修饰语。基本上，循环从999开始，到100结束，然后通过将每个数字修改为负1而到达那里。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3143317|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

The negative one is the modifier. Basically the loops start at 999, end at 100, and get there by modifying each number by negative one.

blocks|key|2186582|text|for+x+in+range(100,+1000):
++++for+y+in+range(100,+1000):
++++++++product+=+x*y
++++++++if+product+>+x*y:
++++++++++++break
++++++++if+str(product)+==+str(product)[::-1]+and+product+>+900000:
++++++++++++print+(x,+y,+product)|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|2186583|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>for x in range(100, 1000):
 for y in range(100, 1000):
 product = x*y
 if product &gt; x*y:
 break
 if str(product) == str(product)[::-1] and product &gt; 900000:
 print (x, y, product)
</code></pre>

<pre><code>n = 0
for a in xrange(999, 100, -1):
 for b in xrange(a, 100, -1):
 x = a * b
 if x &gt; n:
 s = str(a * b)
 if s == s[::-1]:
 n = a * b
print n
</code></pre>

I have a question about this solution to <a href="http://projecteuler.net/index.php?section=problems&amp;id=4" rel="nofollow">the problem</a>.
I know it is right but I am wondering why in the <code>xrange(999,100,-1)</code> the <code>-1</code> is there
For the a and b for loops. Please explain. I am new to this :)

Project Euler problem number 4

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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n = 0for a in xrange(999, 100, -1):    for b in xrange(a, 100, -1):        x = a * b        if x > n:            s = str(a * b)            if s == s[::-1]:                 n = a * bprint n关于的这个解决方案，我有

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