[Solved] All possible set of digits that sum==n in python list [closed]

by 


l=[0,0,0,0,0,0,0,0,0,0,0,0]
n = 11

for i in range (0,n):
        l[i] = 1 

print l
carryidx = 0 
for i in reversed(range(0,n - 1)):
        v = l[ n - 1] + l[i]
        if v > 9:
                while l[carryidx] > 9:
                        carryidx += 1
                l[carryidx] += 1
        else:
                l[n - 1] = v 
        l[i] = l[i] - 1 
        print l

Output

[1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0]
[1, 1, 1, 1, 0, 2, 0, 0, 0, 0, 0, 0]
[1, 1, 1, 0, 0, 3, 0, 0, 0, 0, 0, 0]
[1, 1, 0, 0, 0, 4, 0, 0, 0, 0, 0, 0]
[1, 0, 0, 0, 0, 5, 0, 0, 0, 0, 0, 0]
[0, 0, 0, 0, 0, 6, 0, 0, 0, 0, 0, 0]

3

solved All possible set of digits that sum==n in python list [closed]