blocks|key|4345847|text|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。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4345848|def+centuryFromYear(year):
+return+math.ceil(year/100)+|code-block|syntax|javascript|4345849|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

You can use the ceiling function available in 'math' module to get the desired solution.

<pre><code>def centuryFromYear(year):
 return math.ceil(year/100) 
</code></pre>

blocks|key|4345920|text|使用整数除法，在2000和2017中都能正常工作：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4345921|1+%2B+(year+-+1)+//+100++|code-block|syntax|javascript|4345922|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

With integer division, works properly both for 2000 and for 2017:

<pre><code>1 + (year - 1) // 100 
</code></pre>

blocks|key|4345956|text|a+=+int(input('Find+the+Century+=+'))+|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|4345957|将这个数字除以100|unstyled|4345958|century+=+a+//+100+|4345959|检查年份是否属于同一世纪|4345960|if(a%25100+!=+0):
++++++century+=+century+%2B+1
print(century)|4345961|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|P|8|@]|9|@]|A|$]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$B|C]]|$1|I|3|J|5|F|7|R|8|@]|9|@]|A|$]]|$1|K|3|L|5|6|7|S|8|@]|9|@]|A|$B|C]]|$1|M|3|-4|5|F|7|T|8|@]|9|@]|A|$]]]|N|$]]

<pre><code>a = int(input('Find the Century = ')) 
</code></pre>

Divide the number by 100

<pre><code>century = a // 100 
</code></pre>

Check if the year belongs the same century

<pre><code>if(a%100 != 0):
 century = century + 1
print(century)
</code></pre>

blocks|key|4346047|text|另一种选择是0-9999，它更适合于您的尝试。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4346048|year+=+2018
cent+=+int(str(year).zfill(4)[:2])%2B1
print(cent)|code-block|syntax|javascript|4346049|返回：|4346050|21|4346051|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

Another alternative which works for 0-9999 which is more in the lines of your attempts.

<pre><code>year = 2018
cent = int(str(year).zfill(4)[:2])+1
print(cent)
</code></pre>

Returns:

<pre><code>21
</code></pre>

blocks|key|4345889|text|def+centuryFromYear(year):
++++return+-(-year+//+100)|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|4345890|这是相当古老的，但这是正确的世纪输出。这是负楼层划分|unstyled|4345891|1700+//+100+=+17+1701+//+100+=+17+-+(-1701+//+100)+=+18|4345892|它将楼层分割设为-1701//100，即-18|4345893|适用于所有年份，仅为1行|4345894|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|P|8|@]|9|@]|A|$]]|$1|G|3|H|5|F|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|F|7|R|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|S|8|@]|9|@]|A|$]]|$1|M|3|-4|5|F|7|T|8|@]|9|@]|A|$]]]|N|$]]

<pre><code>def centuryFromYear(year):
 return -(-year // 100)
</code></pre>

its rather old, but this is the correct century output. its a negative floor division

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

it makes the floor division on -1701//100 which is -18

works for all years and is 1 line only

blocks|key|4345925|text|首先从上下文中的year减去1|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4345926|def+centuryFromYear(year):
++++return+(year+-+1)+//+100+%2B+1|code-block|syntax|javascript|4345927|适用于实现以下示例：|4345928|print(centuryFromYear(2000))++#+-->+20
print(centuryFromYear(2001))++#+-->+21
print(centuryFromYear(2017))++#+-->+21|4345929|entityMap^0|8|4|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|T|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|U|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|V|8|@]|D|@]|E|$I|J]]|$1|O|3|-4|5|6|7|W|8|@]|D|@]|E|$]]]|P|$]]

First start by subtracting 1 from the <code>year</code> in context

<pre><code>def centuryFromYear(year):
 return (year - 1) // 100 + 1
</code></pre>

Works for implementing the following examples:

<pre><code>print(centuryFromYear(2000)) # --&gt; 20
print(centuryFromYear(2001)) # --&gt; 21
print(centuryFromYear(2017)) # --&gt; 21
</code></pre>

blocks|key|4345990|text|这对我很有效：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4345991|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+%2B=+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+%2B=+str(i)
++++++++elif+len(xlist)+==+2:
++++++++++++xlist.insert(0,+0)
++++++++++++xlist.insert(1,+0)
++++++++++++x+=+""
++++++++++++for+i+in+xlist:
++++++++++++++++x+%2B=+str(i)
++++++++elif+len(xlist)+==+1:
++++++++++++
++++++++++++xlist.insert(0,+0)
++++++++++++xlist.insert(1,+0)
++++++++++++xlist.insert(2,+0)
++++++++++++x+=+""
++++++++++++for+i+in+xlist:
++++++++++++++++x+%2B=+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+%2B=+str(xlist[k])
++++++++#finally+form+the+century+by+adding+1+to+j+and+calling+it+c.
++++c+=+int(j)+%2B+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+%2B=+str(i)
++++++++return(str(x)+%2B+"+is+en+the+century+"+%2B+str(c)+%2B+"+B.C.")
++++else:
++++++++return(str(x)+%2B+"+is+en+the+century+"+%2B+str(c)+%2B+"+A.C.")|code-block|syntax|javascript|4345992|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

THIS WORKED FOR ME:
<pre><code>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 &quot;minus&quot; sign is in x, set the boolean to true and remove the &quot;minus&quot; for easier handling of the variable
 #(the minus doesn't tell us anything anymore because we already set the boolean)
 negative = False
 if &quot;-&quot; in xlist:
 negative = True
 xlist.remove(&quot;-&quot;)
 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) &gt;= 4:
 pass
 
 
 else:
 if len(xlist) == 3:
 xlist.insert(0, 0)
 x = &quot;&quot;
 for i in xlist:
 x += str(i)
 elif len(xlist) == 2:
 xlist.insert(0, 0)
 xlist.insert(1, 0)
 x = &quot;&quot;
 for i in xlist:
 x += str(i)
 elif len(xlist) == 1:
 
 xlist.insert(0, 0)
 xlist.insert(1, 0)
 xlist.insert(2, 0)
 x = &quot;&quot;
 for i in xlist:
 x += str(i)
 

 n = len(xlist) - 2
 #j is the number formed by the first n characters.
 j = &quot;&quot;
 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 &quot;-&quot; and &quot;B.C.&quot; if negative is true, and &quot;A.C.&quot; if negative is false.
 if negative:
 xlist.insert(0, &quot;-&quot;)
 x = &quot;&quot;
 for i in xlist:
 x += str(i)
 return(str(x) + &quot; is en the century &quot; + str(c) + &quot; B.C.&quot;)
 else:
 return(str(x) + &quot; is en the century &quot; + str(c) + &quot; A.C.&quot;)
</code></pre>

blocks|key|4345936|text|year=+int(input())
century+=+(year+-+1)+//+100+%2B+1
print(century)|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|4345937|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>year= int(input())
century = (year - 1) // 100 + 1
print(century)
</code></pre>

blocks|key|4346243|text|实际上，我有一个最优雅的代码，我将与你分享C版本。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4346244|#include<math.h>
#include<stdio.h>

int+main()+{
++float+x;
++int+y;

++fscanf(stdin,+"%25f",+&x);
++x+=+x+/+100;
++y+=+ceil(x);

++fprintf(stdout,+"Century+%25d+",+y);

++return+0;
}|code-block|syntax|javascript|4346245|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

I actually have one of the most elegant codes for this and I will share the C version with you.
<pre class="lang-c prettyprint-override"><code>#include&lt;math.h&gt;
#include&lt;stdio.h&gt;

int main() {
 float x;
 int y;

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

 fprintf(stdout, &quot;Century %d &quot;, y);

 return 0;
}
</code></pre>

blocks|key|4345982|text|这是正确的答案：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4345983|def+centuryFromYear(year):
++if+year+%25+100+==+0:
++++return+year+//+100+
++else:
++++return+year+//+100+%2B+1|code-block|syntax|javascript|4345984|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

This is the correct answer: 

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

Problem: Given a year, return the century it is in. The first century spans from the year 1 up to and including the year 100, the second - from the year 101 up to and including the year 200, etc.

My Code:

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

 if decimal &gt; 0:
 return integer + 1
 else:
 return integer

print(centuryFromYear(2017))
</code></pre>

This doesn't seem to work in certain cases. like when year = 2001 or year = 2000.

Would anyone be able to provide a more simple piece of code?

Year to Century Function

问题:给定一年，返回它所在的世纪。第一个世纪是从公元1年到公元100年，第二个世纪是从公元101年到公元200年，等等。我的代码：def centuryFromYear(year):    century = year/100     decimal = int(str(century[-2:-1]))    integer = int(str(century)[:2])    if decima

问从年份到世纪的函数
EN

回答 10

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