回文数 I

101位自复制数

阶乘

<pre><code>res = 1 for i in range(1,N+1):     res *= i while not res % 10:     res //= 10 print(str(res)[-K])</code></pre>

好像算法复杂度要求不高，迭代就行

&#x27;&#x27;&#x27;import math def f(N,K):     x=math.factorial(N)     while  x%10==0:         x=x//10     x=str(x)     #n=x[-K]     return x[-K] print(f(N,K)) &#x27;&#x27;&#x27; import math def f(N,K):     x=math.factorial(N)     K=K-1     x=list(str(x))[::-1] while  x[0]==&#x27;0&#x27;:         x=x[1::] return x[K] print(f(N,K))

使用内置的阶乘函数即可

result = 1for i in range (1,N+1): result *=is = str(result).strip(&#x27;0&#x27;)print(s[-K])

将结果转换为字符串，去掉字符串末尾的0

def solve_it(n,k): tmp = 1 for i in range(2,n+1): tmp *= i tmp = int(str(tmp).rstrip(&#x27;0&#x27;)[-(k+2):]) return tmp // 10**(k-1) % 10 print(solve_it(N,K))

取末几位算就行

<pre><code>sum = 1 for i in range(1, N+1): sum = sum * i while sum % 10 == 0: sum = sum // 10 print(str(sum)[-K])</code></pre>

通过

<pre><code>import math y = math.factorial(N) while True: if y % 10 == 0:    y = y/10 else: break z = str(y) print(z[len(z)-K])</code></pre>

才发现Python自带阶乘函数

def f(n): &nbsp; &nbsp; s=1 &nbsp; &nbsp; for i in range(1,n+1): &nbsp; &nbsp; &nbsp; &nbsp;s=(s%100000)*i &nbsp; &nbsp; &nbsp; &nbsp;while s%10==0: &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;s=s/10 &nbsp; &nbsp; return s &nbsp; &nbsp; print str(f(N))[::-1][K-1]

比前几题简单好多

sum=1 if N==0: &nbsp; &nbsp; print(1) else: &nbsp; &nbsp; for i in range(N,0,-1): &nbsp; &nbsp; &nbsp; &nbsp; sum=sum*i &nbsp; &nbsp; L=list(str(sum)) &nbsp; &nbsp; while L[-1]=='0': &nbsp; &nbsp; &nbsp; &nbsp; L.pop() &nbsp; &nbsp; print(int(L[-K]))

简洁明了

<pre class="prettyprint lang-py linenums">def jc(N, K):

 t = 1
 for i in range(1, N + 1):
 t *= i
 T = str(t)
 for j in range(1,len(T)):
 if T[-1] == '0':
 T = T[:-1]
 else:
 break

 return int(T[-K])

print(jc(N,K))</pre>

解题报告六十六

<pre style="padding: 9.5px; font-family: Monaco, Menlo, Consolas, &quot;Courier New&quot;, monospace; font-size: 13px; color: rgb(51, 51, 51); border-radius: 4px; margin-top: 0px; margin-bottom: 10px; margin-left: 0px; line-height: 20px; word-break: break-all; overflow-wrap: break-word; white-space: pre-wrap; background-color: rgb(245, 245, 245); border-width: 1px; border-top-style: solid; border-right-style: solid; border-bottom-style: solid; border-color: rgba(0, 0, 0, 0.15); border-image: initial;">def f(n):
 return reduce(lambda x,y:x*y,range(1,n+1))
def F(n,k):
 temp=str(f(n)).rstrip('0')
 return temp[-k]
print F(N,K)</pre>

探索python特性还是挺重要的

sum=1if N==0:    print(1)else:    for i in range(N,0,-1):        sum=sum*i    L=list(str(sum))    while L[-1]==&#x27;0&#x27;:        L.pop()    print(int(L[-K]))

通过了

你定义的函数没有调用，定义完之后需要调用一下

<pre style="padding: 9.5px; font-family: Monaco, Menlo, Consolas, 'Courier New', monospace; font-size: 13px; color: rgb(51, 51, 51); border-top-left-radius: 4px; border-top-right-radius: 4px; border-bottom-right-radius: 4px; border-bottom-left-radius: 4px; margin-top: 0px; margin-bottom: 10px; word-break: break-all; word-wrap: break-word; white-space: pre-wrap; border: 1px solid rgba(0, 0, 0, 0.14902); background-color: rgb(245, 245, 245);">挑战python栏目的所有题目，题目中所给变量使用前不用声明，也不用赋值，系统自动赋值。</pre>

import mathdef fact(n, k):    a = str(math.factorial(n))    length = len(a)    for i in range(length-1, -1, -1):        if a[i] is '0':            pass        else:            break    print i, k    print a    print a[i - k + 1]    fact(10, 3)

第68题：阶乘

放棋子

最小公倍数

101位自复制数

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