Offer Penalty for prove safety (31): 1 integer number appears (the number of occurrences n is an integer from 1 to 1)

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First, the primer

This series is my brush "to prove safety Offer" brush off the cattle in question notes online, it aims to enhance the ability under its own algorithm.
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Second, the title

Calculated the number of integer of 1 to 13 1 arise, and calculates the number of times an integer of 1 to 1 in 1300 appearing? To this end he especially counted about 1 to 13 contains the digits 1 has 1,10,11,12,13 were therefore appears five times, but for the problem behind him Meizhe. ACMer hope you will help him, and the problem is more generalized, the number of non-negative integer in the range 1 appears (the number of occurrences from 1 to n. 1) can quickly find any.

1, ideas

Method 1: The most intuitive, for each integer of 1 ~ n, respectively, is determined in a number n, in particular, see "the offer to prove safety." Time complexity of this method is O (N * logN), when N is large, the general time out.

Method 2: This category of problems, if not solved intuitive, then usually conducted to find the law, transformed into a mathematical problem. This question is at the "Beauty of Programming" has a more detailed description, the following examples in conjunction with a specific analysis:

Prior to the analysis, we first need to know a rule:

From 1-10, in their single digits, the number 1 appears 1 times.
From 1 to 100, in their ten digits, the digital 1 appears 10 times.
From 1-1000, in their hundreds digit, the number 1 appears 100 times.
And so on, from 1 to 10i, in the second place from the right thereof, the number 1 appears 10 ^ (i - 1) times.

For n = 2134, to find this number from 1 to 2134 2134 1 all numbers. We can be bit-wise analysis of 2134:

(1) in bits, from 1 to 2130, contains 10 213, 213 so the number 1 appears twice, the remaining number of digits only 2131,2132,2133,2134 2131 resin containing a word, left not included under. Therefore, the total number of digits in the number 1 is 213 + 1 = 214.

(2) in ten, from 1 to 2100, contains 21 100, the digital 1 appears 21 * 10 = 210 times, and the remaining ten numbers from 2101 ~ 2134, 2110 ~ 2119 which is only 10 digits in digital bit is 1, the total number of digits in the ten 1 is 210 + 10 = 220.

(3) at the one hundred, from 1 to 2000, it contains two 1000, the digital 1 appears 2 * 100 = 200, the remaining number from 2001 to 2134, from 2100 to 2134 only numbers in the 35 one hundred number is 1, the total number of the hundreds digit numbers 1 to 200 + 35 = 235.

(4) on the one thousand, it contains 0 10000, so the number 1 appeared in 1000 * 0 = 0, one thousand numbers the rest of the numbers 1000 to 1999, only the 1000 figures is 1, so one thousand figures on the total number of 1 bits is 1000.

Therefore, from 1 to 2134 in the n numbers, the total number of the number that appears is 214 + 220 + 235 + 1000 = 1669.

What steps summarized above, you can get such a rule:

For the number n, calculating its first i (i from 1 starts, counting from the right) containing a number of digits of the number of bits:

Assumptions made on the i-th bit of x, then

1. If x> 1 then the number 1 is included in the i-th digit of the upper digit :( + 1) * 10 ^ (i-1) (which is the higher digit bit remains from i + 1 to the maximum number of bits consisting of numbers)

2. If x <1, then the number 1 is included in the i-th digit of the upper digit :() * 10 ^ (i-1)

3. If x == 1, then the number of bits comprising the i-th :( 1 is higher digit) * 10 ^ (i-1) + (lower digit 1) (wherein when the lower digit from the i - 1 digits until the first digit of numbers)

2, programming

python

Code implementation:

# -*- coding:utf-8 -*-
class Solution:
    def NumberOf1Between1AndN_Solution(self, n):
        # write code here
        if n < 1:
            return 0
        # mult位数 sumTime出现1的次数和
        mult, sumTimes = 1, 0
        while n//mult:
            div, mod = divmod(n, mult*10)
            curNum, curMod = divmod(mod, mult)
            if curNum > 1:
                sumTimes += div*mult + mult
            elif curNum == 1:
                sumTimes += div*mult + curMod + 1
            else:
                sumTimes += div*mult
            mult = mult*10
        return sumTimes

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