20年前用的是C++，现在学习python，写一个程序，算双色球中奖的。列举所有C（6，33）*16=17721088种可能的号码，从有据可查到现在的1500多期开奖，每个号码能够中多少个一等奖，多少个二等奖到六等奖。
程序写出来了，但是运行时间不能接受，求优化代码

#双色球所有号码历次中奖查询代码

filename="i:/ssq.txt"   #i:/ssq.txt中存储的是双色球历次中奖号码，形式每行如01182630313303，六个红球一个蓝球号码，共1533行

import time
iend=10    # iend调试用，iend=34时为实际情况，共17721088种可能，调试时iend=10，共1344种可能

start=time.time()


with open(filename) as file_object:
    nums = file_object.readlines()


for i0 in range(1,17):
    for i1 in range(1,iend-5):
        for i2 in range(i1+1,iend-4):
            for i3 in range(i2+1,iend-3):
                for i4 in range(i3+1,iend-2):
                    for i5 in range(i4+1,iend-1):
                        for i6 in range(i5+1,iend):

#构建循环框架，i0为蓝球，i1-i6为红球
#在我的破机子上，循环一遍17721088次，每次只计数加1，用时就到了4.698s
#然后是对每一个i0-i6，判断在1533期中，总共中了几个一等奖到六等奖，

                          n=(i1,i2,i3,i4,i5,i6,i0)

                            for s in nums:                                
                                ired,iblue=0,0
                                s1,s2,s3,s4,s5,s6,s0=int(s[0:2]),int(s[2:4]),int(s[4:6]),int(s[6:8]),int(s[8:10]),int(s[10:12]),int(s[12:14])
                                
                                for i in range(0,6):
                                    if(n[i] in (s1,s2,s3,s4,s5,s6)):
                                        ired=ired+1
                                
                                if(n[6] ==s0):
                                    iblue=iblue+1
                                    
                                if((ired,iblue)==(6,1)):
                                    j1=j1+1
                                if((ired,iblue)==(6,0)):
                                    j2=j2+1
                                if((ired,iblue)==(5,1)):
                                    j3=j3+1
                                if((ired,iblue)==(5,0))or((ired,iblue)==(4,1)):
                                    j4=j4+1
                                if((ired,iblue)==(4,0))or((ired,iblue)==(3,1)):
                                    j5=j5+1
                                if((ired,iblue)==(2,1))or((ired,iblue)==(1,1))or((ired,iblue)==(0,1)):
                                    j6=j6+1

                            print(n,'   ',j1,j2,j3,j4,j5,j6)

  

end=time.time()
print("Running time %s seconds"%(end-start))



#调试时1344种情况，不写文件只打印屏幕要近15s，全1700万种可能就得17721088/1344*15/3600约等于55小时。。求优化鶸代码
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