将for循环和if语句的递归函数转换为迭代函数

Question

挑战是找到所有可能的小于N的数字组合,其总和等于N.例如,当N等于:

• 2
  • 1 + 1 - 1路
• 3
  • 2 + 1
  • 1 + 1 + 1 - 2种方式
• 4
  • 3 + 1
  • 2 + 2
  • 2 + 1 + 1
  • 1 + 1 + 1 + 1 - 4种方式

等等...

现在在python中创建它,以了解我起草此代码的模式第一:

N=5
for d in drange(0,N,1):
    if N-d*4>=0:
        for c in drange(0,N,1):
            if N-d*4-c*3>=0:
                for b in drange(0,N,1):
                    if N-d*4-c*3-b*2>=0:
                        for a in drange(0,N,1):
                            if N-d*4-c*3-b*2-a*1==0:
                                if sum([d,c,b,a])!=1:
                                    print d,c,b,a                            
                    else: break
            else:break
    else:break

1. 然后我将代码更改为此处,其中N = 6及以下:

N=6
for e in drange(0,N,1):
    if N-e*5>=0:
        C0 = N-e*5
        for d in drange(0,N,1):
            if C0-d*4>=0:
                C1 = C0-d*4
                for c in drange(0,N,1):
                    if C1-c*3>=0:
                        C2 = C1-c*3
                        for b in drange(0,N,1):
                            if C2-b*2>=0:
                                C3 = C2-b*2
                                for a in drange(0,N,1):
                                    if C3-a*1==0:
                                        if sum([e,d,c,b,a])!=1:
                                            print e,d,c,b,a                            
                            else: break
                    else:break
            else:break
    else:break

1. 下一版本包含数组以跟踪数字并节省计算空间:

N=6
Nums = drange2(6-1,-1,-1)
Vals = [0]*6
Vars = [0]*6
for Vars[0] in drange(0,N,1):
     if N-Vars[0]*Nums[0]>=0:
         Vals[0] = N-Vars[0]*Nums[0]
         for Vars[1] in drange(0,N,1):
             if Vals[0]-Vars[1]*Nums[1]>=0:
                 Vals[1] = Vals[0]-Vars[1]*Nums[1]
                 for Vars[2] in drange(0,N,1):
                     if Vals[1]-Vars[2]*Nums[2]>=0:
                         Vals[2] = Vals[1]-Vars[2]*Nums[2]
                         for Vars[3] in drange(0,N,1):
                             if Vals[2]-Vars[3]*Nums[3]>=0:
                                 Vals[3] = Vals[2]-Vars[3]*Nums[3]
                                 for Vars[4] in drange(0,N,1):
                                     if Vals[3]-Vars[4]*Nums[4]==0:
                                         if sum([Vars[0],Vars[1],Vars[2],Vars[3],Vars[4]])!=1:
                                             print Vars                           
                             else: break
                     else:break
             else:break
     else:break

1. 然后我想让这个代码在N为100的情况下起作用,我把它递归...

N=48
Nums = drange2(N-1,-1,-1)
Vals = [0]*N
Vars = [0]*(N-1)
count=0
def sumCombos(Number,i):
    if i==0:
        global count        
        for Vars[i] in xrange(0,i+2,1):
            z = Number-Vars[i]*Nums[i]
            if z>=0:
                Vals[i] = z
                sumCombos(Number,i+1)
            else: break
    elif i<Number-2:
        for Vars[i] in xrange(0,i+1,1):
            z = Vals[i-1]-Vars[i]*Nums[i]
            if z >=0:
                Vals[i]=z
                sumCombos(Number,i+1)
            else: break
    elif i==Number-2:
        for Vars[i] in xrange(0,i+3,1):
            if Vals[i-1]-Vars[i]*Nums[i]==0:
                count+=1
sumCombos(N,0)
print count

1. 问题:由于1000000+方法调用需要花费太多时间,所以有没有一种方法可以让我在创建之前的级联效果的情况下进行迭代,而无需输入所有内容？我搜索了网站和其他人如何制作一个涉及for循环和if语句迭代的递归函数,但没有运气这个特定的.请提供任何智慧 - 沙哈3

Answer 1

你为什么要它递归？

>>> from itertools import chain, combinations_with_replacement
>>> n = 7
>>> [i for i in chain.from_iterable(
       combinations_with_replacement(range(1, n), k)
       for k in range(2, n+1))
     if sum(i) == n]

[(1, 6), (2, 5), (3, 4), (1, 1, 5), (1, 2, 4), (1, 3, 3), (2, 2, 3), (1, 1, 1, 4), (1, 1, 2, 3), (1, 2, 2, 2), (1, 1, 1, 1, 3), (1, 1, 1, 2, 2), (1, 1, 1, 1, 1, 2), (1, 1, 1, 1, 1, 1, 1)]

这个问题随着n而增长!所以,大数字需要花费很多时间.