最小子串

月饼节快乐

特殊回文数

for i in range(10000,1000000):     a=str(i)     b=a[::-1] if a==b:         cnt=0         ls=len(a) for c in range(ls):             cnt=cnt+int(a[c]) if cnt==n: print(i)

廉颇老矣，尚能饭否

<pre><code>L_5 = [] L_6 = [] for i in range(1, 10): for j in range(10): for k in range(10): if 2 * i + 2 * j + k == n:                 L_5.append(i * 10001 + j * 1010 + k * 100) if 2 * i + 2 * j + 2 * k == n:                 L_6.append(i * 100001 + j * 10010 + k * 1100) for i in L_5 + L_6: print(i)</code></pre>

两个数组记录，不用排序了

<pre><code>def find_2of5(h): L = [] for i in range(1, 10): if 0 &lt;= h - i &lt;= 9: L.append(str(i) + str(h - i)) return L def find_3of6(h): L = [] for i in range(1, 10): for j in range(0, 10): if 0 &lt;= h - i - j &lt;= 9: L.append(str(i) + str(j) + str(h - i - j)) return L R = [] if n % 2 == 1: for i in (1, 3, 5, 7, 9): L = find_2of5((n - i) // 2) if L: for j in L: R.append(j + str(i) + j[::-1]) else: for i in (0, 2, 4, 6, 8): L = find_2of5((n - i) // 2) if L: for j in L: R.append(j + str(i) + j[::-1]) L = find_3of6(n // 2) if L: for j in L: R.append(j + j[::-1]) R = list(map(int, R)) R.sort() for i in R: print(i)</code></pre>

写它一大片儿

b = [] for i in range(10000, 1000000):     c = [] if str(i) == str(i)[-1::-1]: for j in str(i):             c.append(int(j)) if sum(c) == n:             b.append(i) for i in b: print(i)

运行有点慢，

a=[] for x in range(1,10): &nbsp;&nbsp;&nbsp; for y in range(10): &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; for z in range(10): &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; if x*2+y*2+z==n: &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; a.append(x*10001+y*1010+z*100) &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; if x*2+y*2+z*2==n: &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; a.append(x*100001+y*10010+z*1100) a.sort() for i in a: &nbsp;&nbsp;&nbsp; print i

同时构造5位和6位回文数

首先分析一下，如果n是奇数的话，可能的构成只有是5位的情况，那么只要针对第三位进行下分析就可以了，因为另外四位是对称的，所以第三位只可能是奇数。然后另外四位只要求出两位就可以了。有两位的和是一定的；如果n是偶数的话，可能是5位（中间那位是偶数），也可能是六位代码如下&nbsp;<pre class="prettyprint lang-py linenums">def get_valid_two(num):
	pairs = []
	for first in range(1,10):
		second = num - first
		if second &gt;= 0 and second &lt; 10:
			pairs.append(str(first) + str(second))
	return pairs

def get_valid_three(num):
	pairs = []
	for first in range(1, 10):
		for second in range(10):
			third = num - first - second
			if third &gt;= 0 and third &lt; 10:
				pairs.append(str(first) + str(second) + str(third))
	return pairs

def six_digit(n):
	rsl = []
	n_half = n / 2
	pairs = get_valid_three(n_half)
	for pair in pairs:
		rsl.append(pair+pair[::-1])

	return rsl

def five_digit(n):
	rsl = []
	if n &amp; 1:
		steps = range(1,10,2)
	else:
		steps = range(0,10,2)
	for i in steps:
		remain = n - i
		remain_half = remain / 2
		pairs = get_valid_two(remain_half)
		for pair in pairs:
			rsl.append(pair+str(i)+pair[::-1])
	rsl.sort()
	return rsl

def print_rsl(L):
	for value in L:
		print value

#n = 24
if n &amp; 1:
	print_rsl(five_digit(n))
else:
	print_rsl(five_digit(n))
	print_rsl(six_digit(n))
</pre>

直接构造

l = [] if n &lt;= 45: &nbsp; &nbsp; for i in range(0 if n &lt;= 36 else n - 36, 10): &nbsp; &nbsp; &nbsp; &nbsp; if (n - i) % 2 == 1: &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; # 不可能的情况 &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; pass &nbsp; &nbsp; &nbsp; &nbsp; else: &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; num = (n - i) / 2 &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; for j in range(0, 10): &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; if max(j, num - j) &lt;= 9 and min(j,num-j)&gt;=0: &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; if j!=0: &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; s = str(j) + str(int(num - j)) + str(i) + str(int(num - j)) + str(j) &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; l.append(s) for i in range(0 if n&lt;=36 else int((n-36)/2),10): &nbsp; &nbsp; if (n-2*i)%2==1: &nbsp; &nbsp; &nbsp; &nbsp; pass &nbsp; &nbsp; else: &nbsp; &nbsp; &nbsp; &nbsp; num=(n-2*i)/2 &nbsp; &nbsp; &nbsp; &nbsp; for j in range(0,10): &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; if max(j,num-j)&lt;=9 and min(j,num-j)&gt;=0: &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; if j!=0: &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; s = str(j) + str(int(num - j)) + str(i) + str(i)+str(int(num - j)) + str(j) &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; l.append(s) for i in range(0, len(l)): &nbsp; &nbsp; l[i] = int(l[i]) l.sort() for i in l: &nbsp; &nbsp; print(i)

不用暴力搜索，直接从数字的构造入手就可以了

#! /usr/bin/env python # coding:utf-8 def GetCount(num): &nbsp; &nbsp; res = 0 &nbsp; &nbsp; while num: &nbsp; &nbsp; &nbsp; &nbsp; res += 1 &nbsp; &nbsp; &nbsp; &nbsp; num /= 10 &nbsp; &nbsp; return res def IsPalindrome(num): &nbsp; &nbsp; count = GetCount(num) &nbsp; &nbsp; i = 0 &nbsp; &nbsp; L=[] &nbsp; &nbsp; while i &lt; count: &nbsp; &nbsp; &nbsp; &nbsp; L.append(num%10) &nbsp; &nbsp; &nbsp; &nbsp; num /= 10 &nbsp; &nbsp; &nbsp; &nbsp; i += 1 &nbsp; &nbsp; size = len(L) &nbsp; &nbsp; j = 0 &nbsp; &nbsp; while j &lt; size/2: &nbsp; &nbsp; &nbsp; &nbsp; if L[j] != L[size-1-j]: &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; return False &nbsp; &nbsp; &nbsp; &nbsp; j += 1 &nbsp; &nbsp; else: &nbsp; &nbsp; &nbsp; &nbsp; return True def GetNum(num,n): &nbsp; &nbsp; count = GetCount(num) &nbsp; &nbsp; sum = 0 &nbsp; &nbsp; i = 0 &nbsp; &nbsp; if IsPalindrome(num): &nbsp; &nbsp; &nbsp; &nbsp; while i &lt;count: &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; sum += (num%10) &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; num /= 10 &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; i += 1 &nbsp; &nbsp; if sum == n: &nbsp; &nbsp; &nbsp; &nbsp; return True &nbsp; &nbsp; else: &nbsp; &nbsp; &nbsp; &nbsp; return False if __name__ == '__main__': &nbsp; &nbsp; n = input('please input number:') &nbsp; &nbsp; for i in range(10000,1000000): &nbsp; &nbsp; &nbsp; &nbsp; if GetNum(i,n): &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; print i &nbsp; &nbsp;&nbsp;

C语言思想的处理

<pre style="font-family: 'DejaVu Sans Mono'; font-size: 9pt;">def digit_sum(a):
 t = a
 s = 0
 while t &gt; 0:
 s += t % 10
 t /= 10
 return s

for i in range(10, 100):
 t = n - digit_sum(i) * 2
 if 0 &lt;= t &lt;= 9:
 print str(i) + str(t) + str(i)[::-1]

for i in range(100,1000):
 if digit_sum(i) * 2 == n:
 print str(i) + str(i)[::-1]</pre>

果断穷举了

for a in range(1,10): &nbsp; &nbsp; for b in range(10): &nbsp; &nbsp; &nbsp; &nbsp; for c in range(10): &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; if a+b+c+b+a==n: &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; print '{0}{1}{2}{3}{4}'.format(a,b,c,b,a) for a in range(1,10): &nbsp; &nbsp; for b in range(10): &nbsp; &nbsp; &nbsp; &nbsp; for c in range(10): &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; if a+b+c+c+b+a==n: &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; print '{0}{1}{2}{3}{4}{5}'.format(a,b,c,c,b,a)

第142题：特殊回文数

LTC-男人八题之一：连通图

正方形拼接

编辑距离

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