最大整数

阶乘

回文数 I

def solve_it():     step=0     a=M while(a!=int(str(a)[::-1])):         a=a+int(str(a)[::-1])         step+=1 if(step&gt;=8):         step=0 print(step) return &#x27;&#x27; #your answer solve_it()  # 答案需要输出

刷题狗

count = 1M = str(M)for i in range(0,9): M = str(int(M) + int(M[::-1])) if M == M[::-1]: print(count) break else: count+=1else: print(0)

字符串的逆序

<pre><code>def Hui(n): t = str(n) if t == t[::-1]: return 1 return 0 count = 8 flag = 0 while count: if Hui(M): flag = 1 break else: temp = str(M) M = M + int(temp[::-1]) count -= 1 print(8 - count) if flag else print(0)</code></pre>

简单

def func(M, time):     m = str(M)     N = &quot;&quot;.join(m[i] for i in range(len(m)-1, -1, -1)) if time == 8: return 0 elif int(N) == M: return time else:         M = M + int(N)         time += 1 return func(M, time) print(func(M, 0))

刚学到了倒序数字更简便的写法
def func(M, time):
	N = int(str(M)[::-1])
	if time == 8:
		return 0
	elif N == M:
		return time
	else:
		M = M + N
		time += 1
		return func(M, time)

print(func(M, 0))

终于成功使用了一次递归

def huiwen(x):#判断是否为回文数，是则标记为1     x=str(x) if x==x[::-1]:         flag=1 else:         flag=0 return flag def f(M):#得到回文数迭代的次数     count = 0#记录迭代步数 while True:         m=str(M)[::-1]  #逆序         x=int(m)+M  #数+逆序数         flag=huiwen(x) #判断相加是否为回文数，如果是，迭代结束，返回迭代次数；否则继续迭代并更新M的值 if flag==1:             count+=1 break else:             count+=1             M=x continue return count #M=12 c=f(M) if c&lt;=8: print(c) else: print(0)

def solve_it(m): for step in range(1,8): if m == int(str(m)[::-1]): return step-1 else: m += int(str(m)[::-1]) else: return 0 print(solve_it(M))

逆序字符串即可

count = 0 for i in range(0, 9): if M != int(str(M)[::-1]): M = M + int(str(M)[::-1]) count += 1 else: print(count) break else: print(0)

for i in range(0, 9):
    if M != int(str(M)[::-1]):
        M = M + int(str(M)[::-1])
    else:
        print(i)
        break
else:
    print(0)

没有跳出循环则打印0

#M=87 i=0 while i&lt;=8: &nbsp; &nbsp; if str(M)==str(M)[::-1]: &nbsp; &nbsp; &nbsp; &nbsp; print i &nbsp; &nbsp; &nbsp; &nbsp; break &nbsp; &nbsp; else: &nbsp; &nbsp; &nbsp; &nbsp; M=M+int(str(M)[::-1]) &nbsp; &nbsp; &nbsp; &nbsp; i=i+1 if i==9: &nbsp; &nbsp; print 0

简单题，基本照题目意思翻译就行

import copy i=0 while i&lt;=8: &nbsp; &nbsp; N=list(str(M)) &nbsp; &nbsp; N.reverse() &nbsp; &nbsp; k=int(''.join(N)) &nbsp; &nbsp; j=M+k &nbsp; &nbsp; a=list(str(j)) &nbsp; &nbsp; b=copy.deepcopy(a) &nbsp; &nbsp; b.reverse() &nbsp; &nbsp; i=i+1 &nbsp; &nbsp; if a==b: &nbsp; &nbsp; &nbsp; &nbsp; print(i) &nbsp; &nbsp; &nbsp; &nbsp; break &nbsp; &nbsp; M=j if i&gt;8: &nbsp; &nbsp; print(0)

思路简单

<pre class="prettyprint lang-py linenums">def func(num):

 j = 1
 for i in range(8):

 t = num + int(str(num)[::-1])
 if str(t) == str(t)[::-1]:
 break
 else:
 j += 1
 num = t
 if j &gt;= 8:j = 0
 return j


print(func(M))</pre>

回文数

import copyi=0while i&lt;=8:    N=list(str(M))    N.reverse()    k=int(&#x27;&#x27;.join(N))    j=M+k    a=list(str(j))    b=copy.deepcopy(a)    b.reverse()    i=i+1    if a==b:        print(i)        break    M=jif i&gt;8:    print(0)

通过了

<img alt="惊讶" src="http://2.pythontip.sinaapp.com/static/xheditor/xheditor_emot/default/ohmy.gif" /> count &nbsp;= 0;r=M; while r != int(str(r)[::-1]) and count != 8: &nbsp; &nbsp; r = &nbsp;r + int(str(r)[::-1]); &nbsp; &nbsp; count +=1; print [count,0][count==8]

第69题：回文数 I

National Day

回文素数

单身狗

压缩字符串