Question

问题：给定一年，回归它所处的世纪。第一个世纪从第1年开始，包括第100年，第二年 - 从101年到包括200年，等等。 / p>

我的代码：

def centuryFromYear(year):
    century = year/100 
    decimal = int(str(century[-2:-1]))
    integer = int(str(century)[:2])

    if decimal > 0:
        return integer + 1
    else:
        return integer

print(centuryFromYear(2017))

在某些情况下，这似乎不起作用。比如year = 2001或year = 2000。

是否有人能够提供更简单的代码？

Answer 1

你可以在python 3中使用整数除法运算符//：

def centuryFromYear(year):
    return (year) // 100 + 1    # 1 because 2017 is 21st century, and 1989 = 20th century

print(centuryFromYear(2017))  # --> 21

请注意：这不考虑公元前一个世纪，它使用Dec 31st xy99处的截止日期，有时严格定义为Dec 31st xy00
more info here

如果你想在Dec 31st xy00上设置一个更严格的截止值，你可能会想这样做：

def centuryFromYear(year):
    return (year - 1) // 100 + 1    # 1 because 2017 is 21st century, and 1989 = 20th century

print(centuryFromYear(2017))  # --> 21

Answer 2

Python简单单线解决方案和JavaScript单线解决方案
使用javascript中的内置数学函数来获得单行答案
Math.ceil函数始终将数字四舍五入到下一个大数整数或整数。

// Python one-liner solution

def centuryFromYear(year):
    return (year + 99) // 100

// Javascript one-liner solution

function centuryFromYear(year) {

    return Math.ceil(year/100)

}

Answer 3

使用整数除法，适用于2000年和2017年：

1 + (year - 1) // 100

Answer 4

您可以使用“数学”模块中可用的上限功能来获得所需的解决方案。

def centuryFromYear(year):
 return math.ceil(year/100)

Answer 5

clean

将数字除以100

repoB

检查年份是否属于同一世纪

a = int(input('Find the Century = '))

Answer 6

另一种适用于0-9999的替代方案，更符合您的尝试。

year = 2018
cent = int(str(year).zfill(4)[:2])+1
print(cent)

返回：

Answer 7

def centuryFromYear(year):
    return -(-year // 100)

它很老了，但这是正确的世纪输出。它是负的底数分区

1700 // 100 = 17 1701 // 100 = 17 -（-1701 // 100）= 18

它在-1701 // 100上将楼层划分为-18

所有年份都有效，并且只有1行

Answer 8

首先从上下文中的year减去1开始

def centuryFromYear(year):
    return (year - 1) // 100 + 1

实现以下示例的工作：

print(centuryFromYear(2000))  # --> 20
print(centuryFromYear(2001))  # --> 21
print(centuryFromYear(2017))  # --> 21

Answer 9

这对我有用：

def whatCenturyIsX(x):

    #turn our input into a string for modification
    x = str(x)
    #separate the characters of x into a list for further use
    xlist = list(x)
    #set a boolean to contatin negativity or positivity of the number
    #if the "minus" sign is in x, set the boolean to true and remove the "minus" for easier handling of the variable
    #(the minus doesn't tell us anything anymore because we already set the boolean)
    negative = False
    if "-" in xlist:
        negative = True
        xlist.remove("-")
        for i in xlist:
            x += i

    #to define what century year x is in, we are going to take the approach of adding 1 to the first n characters, when N is the number of digits - 2. This is proved. So:
    
    #also, we need the string to be at least 4 characters, so we add 0's if there are less

    if len(xlist) >= 4:
        pass
        
    
    else:
        if len(xlist) == 3:
            xlist.insert(0, 0)
            x = ""
            for i in xlist:
                x += str(i)
        elif len(xlist) == 2:
            xlist.insert(0, 0)
            xlist.insert(1, 0)
            x = ""
            for i in xlist:
                x += str(i)
        elif len(xlist) == 1:
            
            xlist.insert(0, 0)
            xlist.insert(1, 0)
            xlist.insert(2, 0)
            x = ""
            for i in xlist:
                x += str(i)
        

    n = len(xlist) - 2
    #j is the number formed by the first n characters.
    j = ""
    for k in range(0, n):
        #add the first n characters to j
        j += str(xlist[k])
        #finally form the century by adding 1 to j and calling it c.
    c = int(j) + 1



    #for the final return statement, we add a "-" and "B.C." if negative is true, and "A.C." if negative is false.
    if negative:
        xlist.insert(0, "-")
        x = ""
        for i in xlist:
            x += str(i)
        return(str(x) + " is en the century " + str(c) + " B.C.")
    else:
        return(str(x) + " is en the century " + str(c) + " A.C.")

Answer 10

我用 PHP 解决了这个问题。

function centuryFromYear($year) {
    if ($year % 100 == 0){
        return $year/100;
    }
    else {
        return ceil($year/100);
    }
}

注意：-

ceil() 函数将数字向上舍入为最接近的整数。
要将数字向下舍入到最接近的整数，请查看 floor() 函数。
要舍入浮点数，请查看 round() 函数。

Answer 11

year= int(input())
century = (year - 1) // 100 + 1
print(century)

Answer 12

我实际上是最优雅的代码之一，我将与您共享C版本。

#include<math.h>
#include<stdio.h>

int main() {
  float x;
  int y;

  fscanf(stdin, "%f", &x);
  x = x / 100;
  y = ceil(x);

  fprintf(stdout, "Century %d ", y);

  return 0;
}

Answer 13

这是正确答案：

def centuryFromYear(year):
  if year % 100 == 0:
    return year // 100 
  else:
    return year // 100 + 1

年代到世纪的功能

13 个答案: