问题：给定一年，返回它所在的世纪。第一个世纪从第 1 年到第 100 年，第二个世纪从第 101 年到第 200 年，以此类推。
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))
这在某些情况下似乎不起作用。比如年 = 2001 或年 = 2000。
谁能提供一段更简单的代码？
【问题讨论】：
标签： python python-3.x algorithm
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：-
【讨论】：
我实际上有一个最优雅的代码之一，我将与你分享 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;
}
【讨论】：
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.")
【讨论】：
Python 简单单行解决方案 & JavaScript 单行解决方案
在 javascript 中使用内置的数学函数进行单行回答
Math.ceil 函数总是将数字向上舍入到下一个最大值 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)
}
【讨论】：
year= int(input())
century = (year - 1) // 100 + 1
print(century)
【讨论】：
首先从上下文中的year 中减去 1
def centuryFromYear(year):
return (year - 1) // 100 + 1
适用于实现以下示例：
print(centuryFromYear(2000)) # --> 20
print(centuryFromYear(2001)) # --> 21
print(centuryFromYear(2017)) # --> 21
【讨论】：
def centuryFromYear(year):
return -(-year // 100)
它相当古老，但这是正确的世纪输出。这是一个负面的楼层划分
1700 // 100 = 17 1701 // 100 = 17 - (-1701 // 100) = 18
它在 -1701//100 上进行楼层划分，即 -18
适用于所有年份，并且只有 1 行
【讨论】：
您可以使用“数学”模块中提供的上限函数来获得所需的解决方案。
def centuryFromYear(year):
return math.ceil(year/100)
【讨论】：
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：
21
【讨论】：
这是正确答案：
def centuryFromYear(year):
if year % 100 == 0:
return year // 100
else:
return year // 100 + 1
【讨论】：
a = int(input('Find the Century = '))
将数字除以 100
century = a // 100
检查年份是否属于同一世纪
if(a%100 != 0):
century = century + 1
print(century)
【讨论】：
在python 3中可以使用整数除法，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：这不考虑公元前世纪，它使用Dec 31st xy99 的截止日期，有时严格定义为Dec 31st xy00
more info here
如果您想在Dec 31st xy00 上设置更严格的截止值，您可能希望这样做：
def centuryFromYear(year):
return (year - 1) // 100 + 1 # 1 because 2017 is 21st century, and 1989 = 20th century
print(centuryFromYear(2017)) # --> 21
【讨论】：
使用整数除法，2000 年和 2017 年都可以正常工作：
1 + (year - 1) // 100
【讨论】：
Dec 31st xx99