Python小练习-SDUT 2404 Super Prime

http://acm.sdut.edu.cn/sdutoj/problem.php?action=showproblem&problemid=2404

题目没什么好说的，也不难理解，就是超时问题一直没解决，写了三种方法，前两种是7s左右，最后一种4s。

第一种

View Code

 1 #!/usr/bin/python  

 2 #coding=utf-8  

 3 import math  

 4 max = 100000  

 5 prime = [1]*max  

 6 def isprime():  

 7     for m in xrange(0,2):  

 8         prime[m] = 0  

 9     for j in xrange(3,max):  

10         if j%2 == 0:  

11             prime[j] = 0  

12         else:  

13             prime[j] = 1  

14     xn = int(math.sqrt(max*1.0))  

15     for k in xrange(3,xn+1):  

16         if prime[k]:  

17             for l in xrange(2*k,max,k):  

18                 prime[l] = 0  

19 r = 1  

20 s = [1]*max  

21 isprime()  

22 for f in xrange(2,max):  

23     if prime[f] == 1:  

24         s[r]=f  

25         r=r+1  

26 n = input()  

27 for v in xrange(1,n+1):  

28     flag = 0  

29     o=1  

30     a=input()  

31     if prime[a] == 0:  

32         print "Case %d: no"%v  

33         continue  

34     while s[o]<=a/2+1:  

35         p=o  

36         tmp=a  

37         while tmp>0:  

38             tmp = tmp -s[p]  

39             p=p+1  

40         o=o+1  

41         if tmp == 0:  

42             flag=1  

43             break  

44     if flag:  

45         print "Case %d: yes"%v  

46     else:  

47         print "Case %d: no"%v   

第二种

首先用dic存储n范围内的元素，然后依次标记，最后从头扫描一次dic，把合适的元素放在list中返回结果，据说这样求素数比直接用列表存要快，我测试了一下一千万的数据大概跑7s。

View Code

 1 #!/usr/bin/python   

 2 #coding=utf-8  

 3 def prime(n):  

 4     lis = {}  

 5     for i in xrange(2,n+1):  

 6         if not i in lis:  

 7             lis[i] = 1  

 8             k = i*2  

 9             while k<=n:  

10                 lis[k] = 0  

11                 k = k+i  

12     ans = []  

13     for i in lis:  

14         if lis[i] == 1:  

15             ans.append(i)  

16     return ans  

17 pr = prime(100000)  

18 n = input()  

19 for j in xrange(0,n):  

20     flag = 0  

21     o=0  

22     a=input()  

23     if not a in pr:  

24         print "Case %d: no"%j  

25         continue  

26     while pr[o]<=a/2+1:  

27         p=o  

28         tmp=a  

29         while tmp>0:  

30             tmp = tmp-pr[p]  

31             p+=1  

32         o+=1  

33         if tmp == 0:  

34             flag=1  

35             break  

36     if flag:  

37         print "Case %d: yes"%j  

38     else:  

39         print "Case %d: no"%j   

第三种

这个主要是改进了下面的判断，跑到4s，不过还是超时，希望能找到更好的解法。

View Code

 1 #!/usr/bin/python   

 2 #coding=utf-8  

 3 def prime(n):  

 4     lis = {}  

 5     for i in xrange(2,n+1):  

 6         if not i in lis:  

 7             lis[i] = 1  

 8             k = i*2  

 9             while k<=n:  

10                 lis[k] = 0  

11                 k = k+i  

12     ans = []  

13     for i in lis:  

14         if lis[i] == 1:  

15             ans.append(i)  

16     return ans  

17 pr = []  

18 pr = prime(1000001)  

19 n = input()  

20 cntnum = 0  

21 while n:  

22     n=n-1  

23     cnt = 0  

24     issuper=False  

25     j=0  

26     lower=0  

27     sum=0  

28     a = input()  

29     if a in pr:  

30         while True:  

31             if sum < a:  

32                 if sum+pr[j] > a:  

33                     sum-=pr[lower]  

34                     lower+=1  

35                 else:  

36                     sum+=pr[j]  

37                     j+=1  

38                 if lower > j:  

39                     break  

40             elif sum==a:  

41                 if pr[j]!=pr[lower+1]:  

42                     issuper = True   

43                 break  

44         if issuper:  

45             cntnum+=1  

46             print "Case %d: yes"%cntnum  

47         else:  

48             cntnum+=1  

49             print "Case %d: no"%cntnum  

50     else:  

51         cntnum+=1  

52         print "Case %d: no"%cntnum