Step 1: Find the number of digits your answers are likely to fall in. In your
example it is 2 or 3.
Step 2: For a given number size (number of digits)
Step 2a: Pick the possibilities for the first (most significant digit).
Find the min and max number starting with that digit (ascend or descending
order of rest of the digits). If both of them fall into the range:
step 2ai: Count the number of digits starting with that first digit and
update that count
Step 2b: Else if both max and min are out of range, ignore.
Step 2c: Otherwise, add each possible digit as second most significant digit
and repeat the same step
通过案例解决问题：
对于数字大小为2，即__：
0_ : Ignore since it starts with 0
2_ : Minimum=20, Max=25. Both are in range. So update count by 3 (second digit might be 0,3,5)
3_ : Minimum=30, Max=35. Both are in range. So update count by 4 (second digit might be 0,2,3,5)
5_ : Minimum=50, Max=53. Both are in range. So update count by 3 (second digit might be 0,2,3)
对于大小为3:
0__ : Ignore since it starts with 0
2__ : Minimum=200, max=253. Both are in range. Find the number of ways you can choose 2 numbers from a set of {0,0,3,3,5}, and update the count.
3__ : Minimum=300, max=353. Both are in range. Find the number of ways you can choose 2 numbers from a set of {0,0,2,3,5}, and update the count.
5__ : Minimum=500, max=532. Both are out of range. Ignore.
一个更有趣的情况是当最大限制为522时（而不是400）：
5__ : Minimum=500, max=532. Max out of range.
50_: Minimum=500, Max=503. Both in range. Add number of ways you can choose one digit from {0,2,3,5}
52_: Minimum=520, Max=523. Max out of range.
520: In range. Add 1 to count.
522: In range. Add 1 to count.
523: Out of range. Ignore.
53_: Minimum=530, Max=532. Both are out of range. Ignore.
def countComb(currentVal, digSize, maxVal, minVal, remSet):
minPosVal, maxPosVal = calculateMinMax( currentVal, digSize, remSet)
if maxVal>= minPosVal >= minVal and maxVal>= maxPosVal >= minVal
return numberPermutations(remSet,digSize, currentVal)
elif minPosVal< minVal and maxPosVal < minVal or minPosVal> maxVal and maxPosVal > maxVal:
return 0
else:
count=0
for k in unique(remSet):
tmpRemSet = [i for i in remSet]
tmpRemSet.remove(k)
count+= countComb(currentVal+k, digSize, maxVal, minVal, tmpRemSet)
return count
在您的情况下：
countComb('',2,400,20,['0','0','2','3','3','5']) + countComb('',3,400,20,['0','0','2','3','3','5']) 将会给出答案。
def calculateMinMax( currentVal, digSize, remSet):
numRemain = digSize - len(currentVal)
minPosVal = int( sorted(remSet)[:numRemain] )
maxPosVal = int( sorted(remSet,reverse=True)[:numRemain] )
return minPosVal,maxPosVal
numberPermutations(remSet,digSize, currentVal): Basically number of ways
you can choose (digSize-len(currentVal)) values from remSet. See permutations
with repeats.
