Question

问题是：

素数3,7,109和673非常值得注意。通过取任何两个素数并以任何顺序连接它们，结果将是永远是素数。例如，取7和109，7109和1097都是素数。这四个素数的总和，792，代表了具有此属性的一组四个素数的最低总和。找到一组五个素数的最低和，其中任何两个素数连接以产生另一个素数。

现在，以下是未在结果列表中提供任何输出的程序。请帮忙

LIMIT = 20000

prima = []  #empty list

def Bsearch(lsta,low,high,search):   #Binary search function to search a prime number
    if low>high:
        return 0
    else:
        mid = int((low+high)/2)

        if search<lsta[mid]:
            Bsearch(lsta,low,mid-1,search)
        if search>lsta[mid]:
            Bsearch(lsta,mid+1,high,search)
        if search==lsta[mid]:
            return 1
    return 0

def primes(LIMIT):   #sieve to create prime numbers upto LIMIT
    dic = {}    #empty dictionary
    for i in range(2,LIMIT):
        dic[i] = 1
    for i in range(2,LIMIT):
        for j in range(i,LIMIT):
            if i*j>LIMIT:
                break
            dic[i*j] = 0
    for i in range(2,LIMIT):
        if dic[i]==1:
            prima.append(i)

primes(LIMIT)

result = []

for i in range(0,len(prima)):
    print(str((i/len(prima)*100))+"% list passed")

    tempa = []
    tempa.append(prima[i])
    count = 0
    for j in range(i+1,len(prima)):
        c1 = int(str(prima[j])+str(prima[i]))   #first combination
        c2 = int(str(prima[i])+str(prima[j]))   #second combination
        if(Bsearch(prima,0,len(prima)-1,c1) and Bsearch(prima,0,len(prima)-1,c2)):
            print("small success : "+str(count))
            tempa.append(prima[j])
            count +=1
        if(count==4):
            result.append(tempa)
            print("success!")
            break

for item in result:
    print(item)

Answer 1

由于您的搜索是递归的，因此您需要返回搜索结果。

def Bsearch(lsta,low,high,search):   #Binary search function to search a prime number
  retval = 0
  if low>high:
    retval = 0
  else:
    mid = int((low+high)/2)

    if search == lsta[mid]:
        # Make this test first as allows it to exit at once with success
        retval = 1
    elif search<lsta[mid]:
        retval = Bsearch(lsta,low,mid-1,search)
    else:     # search>lsta[mid] Since only 3 choices elif not needed
        retval = Bsearch(lsta,mid+1,high,search)
  return retval

Answer 2

LIMIT = 20000

prima = []  #empty list

def Bsearch(lsta,low,high,search):   #Binary search function to search a prime number
    if low>high:
        return 0
    else:
        mid = int((low+high)/2)

        if search<lsta[mid]:
            Bsearch(lsta,low,mid-1,search)
        if search>lsta[mid]:
            Bsearch(lsta,mid+1,high,search)
        if search==lsta[mid]:
            return 1
    return 0

def primes(LIMIT):   #sieve to create prime numbers upto LIMIT
    dic = {}    #empty dictionary
    for i in range(2,LIMIT):
        dic[i] = 1
    for i in range(2,LIMIT):
        for j in range(i,LIMIT):
            if i*j>LIMIT:
                break
            dic[i*j] = 0
    for i in range(2,LIMIT):
        if dic[i]==1:
            prima.append(i)

primes(LIMIT)

result = []

Answer 3

var jqxhr = $.post(url_update, { id: id , name: name});

Python程序没有输出

3 个答案: