【每日一题20220613】9的时代

[hogwarts (霍格沃兹测试学院官方)]

‍ 今天的主角是数字9。给定一个正整数n，请编写一个函数，统计从数字1到n（包含）中出现了多少个9。例如99，919都算有2个9。

【示例】
输入：10
输出：1
解释：数字1到10中,只有数字9满足条件，所以算作1个。

题目难度：中等
题目来源：codewars-count '9’s from 1 to n

def solution(n: int)-> int:
    # your code here

assert solution(8) == 0
assert solution(10) == 1
assert solution(100) == 20
assert solution(279) == 48

# 进阶
assert solution(99999) == 50000
assert solution(565754) == 275645

[hogwarts_wenzi (蚊子)]

先来个最low的实现

import re


def solution(n: int) -> int:
    # your code here
    count = 0
    for i in range(1,n+1):
        num = len(re.findall("9", str(i)))
        count += num
    return count


assert solution(8) == 0
assert solution(10) == 1
assert solution(100) == 20

[joyoko (徐子宇)]

def solution(n: int) -> int:
    # your code here
    result = 0
    for i in range(1, n + 1):
        count = str(i).count('9')
        result += count
    return result


assert solution(8) == 0
assert solution(10) == 1
assert solution(100) == 20

[xiaopeng (霍格沃兹_小鹏)]

def solution(n: int) -> int:
    # your code here
    str_n = str(n)
    carry_num = 1
    count = 0
    nine_num = 0
    if int(str_n[-1]) >= 9:
        count += 1
    str_n = str_n[:-1]
    while (True):
        if len(str_n) == 0:
            break
        count += int(str_n[-1])*(carry_num*10**(carry_num-1))
        if str_n[-1] == '9':
            nine_num+=int(str(n)[len(str_n):]) + 1
        str_n = str_n[:-1]
        carry_num+=1
    count+= nine_num
    return count

[yoyoyang1004_9968]

def solution(n:int) -> int:
    count = 0
    for i in range(1,n+1):
        count += str(i).count('9')
    return count

assert solution(8) == 0
assert solution(10) == 1
assert solution(100) == 20

[Amoyshmily (霍格沃兹-百里)]

def solution(n: int) -> int:
    lens = len(str(n))  # 目标数字长度

    if lens == 1:  
        return 1 if n == 9 else 0  # 递归结束条件
    
    base = 10 ** (lens - 1)  # 辅助数字
    nines = (lens - 1) * 10 ** (lens - 2)
    
    d, m = divmod(n, base)  # 商数和除数
    value = d * nines + solution(m)  # 递归计算对应数量
    
    if d == 9:
        value += m + 1  # 单独处理9开头的
    return value

assert solution(8) == 0
assert solution(10) == 1
assert solution(100) == 20
# 进阶
assert solution(99999) == 50000
assert solution(565754) == 275645

[Amoyshmily (霍格沃兹-百里)]

解释：

第一步，总结规律，可以推断出：

个位：当为9时1个，其他0个
十位（10）: 1个
百位（100）: 20个
千位（1000）: 300个
万位（10000）: 4000个

也就是说，多位数中，某个位置上包含9的数字个数是： (长度 - 1) * 10 ** (长度 - 2)，备用。

第二步，将目标数字拆解，分别统计每个位置上获取对应含9的数量。
例如279，可以拆分为200+70+9。已知100中含有20个9，那么200就是20*2=40个9；已知每10内有1个9，70就有7个9（09，19，…，69）;最后个位上9算1个，总共就是40+7+1=48个。

可以借用divmod()函数，会同时得到商（整除部分）和余数（除不尽部分）。

另外注意，需要特殊处理下开头本身带9的数字，例如999，每个位置上额外加上对应余数和自身1个即可。比如考虑900的时候，其实只会统计0～899中9的个数，因为首位是9，其他位置任何数字都可以组成含9的组合，因此还要加上余数99个，另外还有首位自身1个。

[zzh]

def solution(n: int)-> int:
count =0
for i in range(1,n+1):
s=list(str(i))
while ‘9’ in s:
s.remove(‘9’)
count =count+1
return count

assert solution(8) == 0
assert solution(10) == 1
assert solution(100) == 20
assert solution(279) == 48
assert solution(99999) == 50000
assert solution(565754) == 275645

[cheng0637007057 (棒棒走天涯)]

def solution(n):
    result = 0
    for i in range(1,n+1):
        result += str(i).count('9')
    return result

[5274080738_8698]

def solution(n: int)-> int:
    cout = 0
    for i in range(n+1):
        cout += str(i).count('9')
    return cout

assert solution(8) == 0
assert solution(10) == 1
assert solution(100) == 20
assert solution(279) == 48

# 进阶
assert solution(99999) == 50000
assert solution(565754) == 275645

[Amoyshmily (霍格沃兹-百里)]

规律：

个位：当为9时1个，其他0个
十位（10）: 1个
百位（100）: 20个
千位（1000）: 300个
万位（10000）: 4000个
......

def solution(n: int) -> int:
    
    s = str(n)  # 转成字符串
    
    total = 0 
    for i in range(len(s)):
        
        base = i*10**(i-1)  # 找规律
        total += int(s[~i])*int(base)  # 倒着取数
        
        if s[i] == "9":
            # 单独处理字面含9的情况
            total += int(s[i+1::] or 0) + 1

    return total
              
assert solution(8) == 0
assert solution(10) == 1
assert solution(100) == 20
assert solution(279) == 48
assert solution(99999) == 50000
assert solution(565754) == 275645

[liuhaixian (会游泳的鱼)]

[Olivia]

def solution(n: int)-> int:
    count = 0
    for i in range(n+1):
        count = count + str(i).count("9")

    return count

assert solution(8) == 0
assert solution(10) == 1
assert solution(100) == 20
assert solution(279) == 48

assert solution(99999) == 50000
assert solution(565754) == 275645

[oo7xa07ur6b5dbrre79o (齐梓深)]

参考大神们的运行的

[joyoko (徐子宇)]

数据量不大的情况下这个解法没毛病，不过数据量大了，百里老师那个推导出来的数学模型才是最优解。【每日一题20220613】9的时代 - #7，来自 Amoyshmily

[manu.chen_7146]

def solution(n:int)->int:
    count_9 = 0
    for i in range(1,n+1):
        count_9 += str(i).count('9')
    return count_9

assert solution(8) == 0
assert solution(10) == 1
assert solution(100) == 20
assert solution(279) == 48

#进阶
assert solution(99999) == 50000
assert solution(565754) == 275645

[longmin]

def solution(n: int)-> int:
    return ''.join([str(i) for i in range(1,n+1)]).count('9')

[455018255_7468]

def solution(n: int)-> int:
    # your code here
    s = 0
    for i in range(1, n+1):
        if "9" in str(i):
            for j in str(i):
                if j == "9":
                    s += 1
    return s

[lekaixin (ALe阿乐)]

def solution(n: int) -> int:
    lenn=len(str(n))
    if lenn==1:
        if n==9:
            return 1
        else:
            return 0
    base=(lenn-1)*10**(lenn-2)
    shang=n//(10**(lenn-1))
    yu=n%(10**(lenn-1))
    res=shang*base+solution(yu)
    if shang==9:
        res+=(yu+1)
    return res

assert solution(8) == 0
assert solution(10) == 1
assert solution(98) == 18
assert solution(100) == 20
assert solution(279) == 48
assert solution(998) == 297
assert solution(999) == 300
assert solution(1000) == 300
assert solution(10000) == 4000

# 进阶
assert solution(99999) == 50000
assert solution(565754) == 275645