反素数(逆向拼写的素数)

计算表达式

回文素数

import math s=0 c=0 def panduan(m): for i in range(2,int(math.sqrt(m))+1): if m%i==0: return False return True for i in range(2,100000): if panduan(i)==True and i==int(str(i)[::-1]):         s=i         c+=1 if c==n: print(s) break

小白题解

<pre><code>def prime(x): if x == 1: return False for i in range(2, int(math.sqrt(x)) + 1): if x % i == 0: return False return True     i = 0     count = 0 while count &lt; n:         i += 1         s = str(i)         s_r = s[::-1] if s == s_r and prime(i):             count += 1 </code></pre>

Easy

c = [] for i in range(2,1000):     a = 1     j = 2 while j &lt; i: if i%j == 0:             a = 0 break else:             j += 1 if a :         c.append(i) b = [] for i in c: if str(i) == str(i)[-1::-1]:         b.append(i) print(b[n-1])

上限1000

L = []M = []#10000以内的素数for i in range(2,10001): j=2 for j in range(2,i): if i%j ==0: break else: L.append(i)#找出回文素数afor a in L: if str(a)[::-1] ==str(a): M.append(a)print (M[n-1])

按部就班

<pre class="prettyprint lang-py linenums">def isPrime(n):
 if n &lt;= 1:
 return False
 i = 2
 while i * i &lt;= n:
 if n % i == 0:
 return False
 i += 1
 return True


L = [2]

for i in range(2, 100000):
 if i % 2 and str(i) == str(i)[::-1]:
 if isPrime(i):
 L.append(i)

print(L[n - 1])</pre>

def isPrime(n):    if n &lt;= 1:        return False    i = 2    while i * i &lt;= n:        if n % i == 0:            return False        i += 1    return TrueL = [2]for i in range(2, 100000):    if i % 2 and str(i) == str(i)[::-1]:        if isPrime(i):            L.append(i)print(L[n - 1])

 反素数

def h(i): &nbsp; &nbsp; if str(i)==str(i)[::-1]: &nbsp; &nbsp; &nbsp; &nbsp; return 1 &nbsp; &nbsp; else: &nbsp; &nbsp; &nbsp; &nbsp; return 0 def z(m): &nbsp; &nbsp; for i in range(2,int(m**0.5+1)): &nbsp; &nbsp; &nbsp; &nbsp; if m%i==0: &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; return 0 &nbsp; &nbsp; else: &nbsp; &nbsp; &nbsp; &nbsp; return 1 def f(n): &nbsp; &nbsp; L=[2,3,5,7,11] &nbsp; &nbsp; if n&lt;=5: &nbsp; &nbsp; &nbsp; &nbsp; return L[:n]&nbsp; &nbsp; &nbsp; else: &nbsp; &nbsp; &nbsp; &nbsp; t=13 &nbsp; &nbsp; &nbsp; &nbsp; while len(L)&lt;n: &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; if h(t) and z(t): &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; L.append(t) &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; t=t+2 &nbsp; &nbsp; &nbsp; &nbsp; return L print f(n)[-1]

放松一下

def zhishu(n): &nbsp; &nbsp; flag=True &nbsp; &nbsp; for i in range(2,n): &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; if n%i==0: &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; flag=False &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; break &nbsp; &nbsp; if flag: &nbsp; &nbsp; &nbsp; &nbsp; return True &nbsp; &nbsp; else: &nbsp; &nbsp; &nbsp; &nbsp; return False &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp; L=[] k=3 if n==1: &nbsp; &nbsp; print(2) else: &nbsp; &nbsp; cout=1 &nbsp; &nbsp; while cout!=n: &nbsp; &nbsp; &nbsp; &nbsp; m=str(k) &nbsp; &nbsp; &nbsp; &nbsp; a=int(m[::-1]) &nbsp; &nbsp; &nbsp; &nbsp; if zhishu(k) and zhishu(a) and a==k: &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; cout+=1 &nbsp; &nbsp; &nbsp; &nbsp; k=k+2 &nbsp; &nbsp; print(k-2)

类似最后一题

def fun(x): &nbsp; &nbsp; flag=False &nbsp; &nbsp; for i in range(2,x): &nbsp; &nbsp; &nbsp; &nbsp; if x%i==0: &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; flag=True &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; break &nbsp; &nbsp; if(flag): &nbsp; &nbsp; &nbsp; &nbsp; return False &nbsp; &nbsp; return True def reverse(x): &nbsp; &nbsp; Str=str(x) &nbsp; &nbsp; Str=Str[::-1] &nbsp; &nbsp; return int(Str) n=100 cnt=0 for i in range(2,100000): &nbsp; &nbsp; if(i==reverse(i)): &nbsp; &nbsp; &nbsp; &nbsp; if(fun(i) and fun(reverse(i))): &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; cnt=cnt+1 &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; if(cnt==n): &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; print(i) &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; break#先判断回文数再判断是否为负数，这样会快一点

简单

num = 0 for i in range(2, 100000): &nbsp; &nbsp; sum = 0 &nbsp; &nbsp; if str(i) == str(i)[::-1]: &nbsp; &nbsp; &nbsp; &nbsp; for j in range(2, i + 1): &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; if not i%j: &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; sum += 1 &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; if sum &gt; 1: &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; break &nbsp; &nbsp; &nbsp; &nbsp; if sum == 1: &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; num += 1 &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; if num == n: &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; print(i) &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; break

笨办法，改变判断条件的顺序，可以很大程度提高执行效率

def zhishu():     count = 2 while True: for i in range(2, count + 1): #不是素数 if count % i == 0 and count != i: break else: #回文素数 if str(count) == str(count)[::-1]: yield count         count += 1 def test(num,func): for i in range(1,num+1): if i == num: print(next(func)) break next(func) a=zhishu() test(n,a)

第177题：回文素数

National Day

特殊回文数

分拆素数和

最大连续子序列