RSA密码方程

平分果子

IP判断

s=s.split(&quot;.&quot;) end = &quot;Yes&quot; if len(s) !=4: end = &quot;No&quot; for i in s: if not i.isdigit(): end = &quot;No&quot; break if len(i)&gt;len(str(int(i))): end = &quot;No&quot; break if int(i) &lt;0 or int(i)&gt;255: end = &quot;No&quot; break print(end)

常规判断

<pre><code>flag = 0 temp = s.split(&#x27;.&#x27;) if len(temp) != 4: #看看是不是四部分组成 flag = 1 for each in temp: #数字组成字符，没有前导零（本身就是0除外），不能大于255，不能小于0 if not each.isdigit() or (&#x27;0&#x27; == each[0] and each != &#x27;0&#x27;) \ or int(each) &gt; 255 or int(each) &lt; 0: flag = 1 break if flag: print(&#x27;No&#x27;) else: print(&#x27;Yes&#x27;) </code></pre>

提交几次了，小问题好多

print(&quot;Yes&quot; if len(list(filter(lambda x:x.isdigit() and 255&gt;=int(x)&gt;=0,s.split(&#x27;.&#x27;))))==4 else &quot;No&quot;)

一行

st=s.split(&quot;.&quot;)if len(st)!=4: #判断是否有四段 print(&quot;No&quot;) sys.exit()for a in st: if len(a)==0 : #判断里面有没有内容 print(&quot;No&quot;) sys.exit() if a[0]==&#x27;0&#x27; and len(a)!=1: #判断是不是0开头的非法数字 print(&quot;No&quot;) sys.exit() if a.isdigit() and 0&lt;=int(a)&lt;=255: #判断是不是数字，数字的范围 pass else: print(&quot;No&quot;) sys.exit()print(&quot;Yes&quot;)

考虑的东西还挺多的

b = 0 for each in s.split(sep = &quot;.&quot;): if each ==&quot;&quot; or (each[0] == &quot;0&quot; and len(each) != 1) or not each.isdigit(): #排除不到四位、开头等于0和不全是数字的情况         b = 0 break if int(each) &gt;= 0 and int(each) &lt; 256:         b +=1 print(&quot;Yes&quot; if b == 4 else &quot;No&quot;)

来个短一点的

def solve_it(s): try: l = s.split(&#x27;.&#x27;) if len(l) != 4: return &quot;No&quot; else:#进行类型转换 res = list(map(int,l)) for i in res: if 0 &lt;= i &lt;= 255: continue else: return &quot;No&quot; else: return &quot;Yes&quot;#转换异常输出No except ValueError: return &quot;No&quot; print(solve_it(s))

强制类型转换

<pre><code>a = s.split(&#x27;.&#x27;) flag = True if len(a) != 4: flag = False else: try: for i in a: if int(i) &lt; 0 or int(i) &gt; 255: flag = False break elif str(int(i)) != i: flag = False break except ValueError: flag = False print(&#x27;Yes&#x27; if flag else &#x27;No&#x27;)</code></pre>

先判断长度，再判断是否开头儿有0

<pre><code>import re if re.findall(&#x27;[1-2]?[0-5]?[0-5]\.[1-2]?[0-5]?[0-5]\.[1-2]?[0-5]?[0-5]\.[1-2]?[0-5]?[0-5]&#x27;, s) == [s]: print(&#x27;Yes&#x27;) else: print(&#x27;No&#x27;)</code></pre>

非常简单

L=s.split('.') flag=1 for i in range(len(L)): &nbsp; &nbsp; if len(L)!=4: &nbsp; &nbsp; &nbsp; &nbsp; flag=0 &nbsp; &nbsp; if L[i].isdigit(): &nbsp; &nbsp; &nbsp; &nbsp; if int(L[i])&lt;0&nbsp; or int(L[i])&gt;255: &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; flag=0 &nbsp; &nbsp; else: &nbsp; &nbsp; &nbsp; &nbsp; flag=0 if flag==1: &nbsp; &nbsp; print('Yes') else: &nbsp; &nbsp; print('No')

c = 0
a = s.split(".")
if len(a) !=4:
 print("No")
else:
 for i in a:
 m = i.isdigit()
 if m == True:
 if int(i)<= 255 and int(i)>= 0:
 c +=1
 else:
 pass
 if c == 4:
 print("Yes")
 else:
 print("No")

初学者代码，简单，易懂

<div style="color: rgb(51, 51, 51); background-color: rgb(250, 250, 250); font-family: Consolas, &quot;Courier New&quot;, monospace; font-size: 18px; line-height: 24px; white-space: pre;"><div>l = s.split('.')</div><div>out = 'Yes'</div><div>if len(l) != 4:</div><div> out = 'No'</div><div>else:</div><div> for i in l:</div><div> if i.isdigit():</div><div> if int(i) == 0:</div><div> continue</div><div> if i.find('0') == 0 or int(i) &lt; 0 or int(i) &gt; 255:</div><div> out = 'No'</div><div> break</div><div> continue</div><div> else:</div><div> out = 'No'</div><div> break </div><div>print(out)</div></div>

考虑0的情况

通过了

<pre style="font-family: 宋体; font-size: 12pt;">flag = []
lst_s = s.split('.')
for ls in lst_s:
 if ls.isdigit():
 if int(ls) &gt;= 0 and int(ls) &lt;= 255:
 if (len(ls) &gt; 1 and ls[0] != '0') or len(ls) == 1:
 flag.append(&quot;True&quot;)
if len(flag) == 4:
 print(&quot;Yes&quot;)
else:
 print(&quot;No&quot;)</pre>

最笨的办法

def f(s): &nbsp; &nbsp; ss = s.split('.') &nbsp; &nbsp; if len(ss) != 4: &nbsp; &nbsp; &nbsp; &nbsp; return 'No' &nbsp; &nbsp; for i in range(4): &nbsp; &nbsp; &nbsp; &nbsp; try: &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; if int(ss[i]) not in range(0,256): &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; return 'No' &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; elif (int(ss[i]) / (10 ** (len(ss[i])-1)) == 0): &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; return 'No' &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; except: &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; return 'No' &nbsp; &nbsp; return 'Yes' print f(s) &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;

题目错了，加了前导0的判断反而错了，去掉就对了

<pre style="font-family: 宋体; font-size: 10.5pt; background-color: rgb(64, 64, 64);">ip首位不能为0 ，我加上这个判断反而错了，去掉反而对了</pre><pre style="font-family: 宋体; font-size: 10.5pt; background-color: rgb(64, 64, 64);">判断首位为0</pre><pre style="font-family: 宋体; font-size: 10.5pt; background-color: rgb(64, 64, 64);">print &quot;No&quot; if len(s.split(&quot;.&quot;)) !=4 or len([i for i in s.split(&quot;.&quot;) if i.isdigit() and str(int(i)) == i]) !=4 or int(s.split(&quot;.&quot;)[0]) == 0 or int(max(s.split(&quot;.&quot;))) &gt;255 or int(min(s.split(&quot;.&quot;))) &lt;0 else &quot;Yes&quot;</pre><pre style="font-family: 宋体; font-size: 10.5pt; background-color: rgb(64, 64, 64);">未判断首位为0</pre><pre style="font-family: 宋体; font-size: 10.5pt; background-color: rgb(64, 64, 64);">print &quot;No&quot; if len(s.split(&quot;.&quot;)) !=4 or len([i for i in s.split(&quot;.&quot;) if i.isdigit() and str(int(i)) == i]) !=4 or int(max(s.split(&quot;.&quot;))) &gt;255 or int(min(s.split(&quot;.&quot;))) &lt;0 else &quot;Yes&quot;</pre><pre style="font-family: 宋体; font-size: 10.5pt; background-color: rgb(64, 64, 64);"><pre style="font-family: 宋体; font-size: 10.5pt; background-color: rgb(64, 64, 64);">
</pre> </pre>

题目有问题

import re import sys s=&quot;255.11.11.11&quot; c=0 b = s.split('.') if len(b)!=4: &nbsp; &nbsp; print 'No' else: &nbsp; &nbsp; for i in range(0,4): &nbsp; &nbsp; &nbsp; &nbsp; if re.match('^[0-9]+$',b[i]): &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; if int(b[i])&gt;=0 and int(b[i])&lt;=255: &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; c=c+1 &nbsp; &nbsp; if c==4: &nbsp; &nbsp; &nbsp; &nbsp; print 'Yes' &nbsp; &nbsp; else: &nbsp; &nbsp; &nbsp; &nbsp; print 'No'

方法比较笨

import re m = re.match(&quot;^([1-9][0-9]{0,2})\.(0|[1-9][0-9]{0,2})\.(0|[1-9][0-9]{0,2})\.(0|[1-9][0-9]{0,2})$&quot;, s) print &quot;Yes&quot; if m and all(int(i)&lt;=255 for i in m.groups()) else &quot;No&quot; ##########错对分界#############import re m = re.match(&quot;^(0|[1-9][0-9]{0,2})\.(0|[1-9][0-9]{0,2})\.(0|[1-9][0-9]{0,2})\.(0|[1-9][0-9]{0,2})$&quot;, s) print &quot;Yes&quot; if m and all(int(i)&lt;=255 for i in m.groups()) else &quot;No&quot;

第64题：IP判断

进制转换

求解幂方程

杨辉三角

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