blocks|key|4494113|text|您应该使用datetime库，它将自动处理所有的闰年信息：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4494114|from+datetime+import+date
from+collections+import+Counter

counter+=+Counter()

for+year+in+xrange(1901,+2001):
++++for+month+in+xrange(1,+13):
++++++++day+=+date(year,+month,+1)
++++++++counter[day.weekday()]+%2B=+1

print+counter[6]|code-block|syntax|javascript|4494115|entityMap^0|5|8|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

You should use the <code>datetime</code> library, which will handled all the leap year information automatically:

<pre><code>from datetime import date
from collections import Counter

counter = Counter()

for year in xrange(1901, 2001):
 for month in xrange(1, 13):
 day = date(year, month, 1)
 counter[day.weekday()] += 1

print counter[6]
</code></pre>

blocks|key|4494180|text|您正在迭代months字典，希望它按月份的顺序迭代，但字典没有排序，因此您可能会以错误的顺序获得月份。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4494181|由于您实际上并不需要月份名称，因此您可以将月份长度列表设置为months。|4494182|entityMap^0|5|6|0|U|6|0^^$0|@$1|2|3|4|5|6|7|J|8|@$9|K|A|L|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|H|3|-4|5|6|7|P|8|@]|D|@]|E|$]]]|I|$]]

You're iterating over the <code>months</code> dict, expecting it to iterate in the order of the months, but dicts aren't ordered, so you can get the months in the wrong order.

Since you don't actually need the month names, you can just make <code>months</code> a list of the month lengths instead.

blocks|key|4493101|text|import+time
from+math+import+floor

"""
Gaussian+algorithm+to+determine+day+of+week
"""
def+day_of_week(year,+month,+day):
++++"""
++++w+=+(d%2Bfloor(2.6*m-0.2)%2By%2Bfloor(y/4)%2Bfloor(c/4)-2*c)+mod+7
++++Y+=+year+-+1+for+January+or+February
++++Y+=+year+for+other+months
++++d+=+day+(1+to+31)
++++m+=+shifted+month+(March+=+1,+February+=+12)
++++y+=+last+two+digits+of+Y
++++c+=+first+two+digits+of+Y
++++w+=+day+of+week+(Sunday+=+0,+Saturday+=+6)
++++"""

d+=+day
m+=+(month+-+3)+%25+12+%2B+1
if+m+&gt;+10:+Y+=+year+-+1
else:+Y+=+year
y+=+Y+%25+100
c+=+(Y+-+(Y+%25+100))+/+100

w+=+(d+%2B+floor(2.6+*+m+-+0.2)+%2B+y+%2B+floor(y/4)+%2B+floor(c/4)+-+2*c)+%25+7

return+int(w)

"""
Compute+the+number+of+months+starting+on+a+given+day+of+the+week+in+a+century
"""
def+months_start_range(day,year_start,year_end):
++++total+=+0
++++for+year+in+range(year_start,+year_end+%2B+1):
++++++++for+month+in+range(1,13):
++++++++++++if+day_of_week(year,+month,+1)+==+day:+total+%2B=+1
++++return+total

start+=+time.time()

total+=+months_start_range(0,1901,2000)

elapsed+=+time.time()+-+start

print("%25s+found+in+%25s+seconds")+%25+(total,elapsed)|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|4493102|这可能会解决这个问题。|unstyled|4493103|它花了大约0.068秒来解决它。|4493104|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|L|8|@]|9|@]|A|$]]|$1|G|3|H|5|F|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|F|7|N|8|@]|9|@]|A|$]]]|J|$]]

<pre><code>import time
from math import floor

"""
Gaussian algorithm to determine day of week
"""
def day_of_week(year, month, day):
 """
 w = (d+floor(2.6*m-0.2)+y+floor(y/4)+floor(c/4)-2*c) mod 7
 Y = year - 1 for January or February
 Y = year for other months
 d = day (1 to 31)
 m = shifted month (March = 1, February = 12)
 y = last two digits of Y
 c = first two digits of Y
 w = day of week (Sunday = 0, Saturday = 6)
 """

d = day
m = (month - 3) % 12 + 1
if m &amp;gt; 10: Y = year - 1
else: Y = year
y = Y % 100
c = (Y - (Y % 100)) / 100

w = (d + floor(2.6 * m - 0.2) + y + floor(y/4) + floor(c/4) - 2*c) % 7

return int(w)

"""
Compute the number of months starting on a given day of the week in a century
"""
def months_start_range(day,year_start,year_end):
 total = 0
 for year in range(year_start, year_end + 1):
 for month in range(1,13):
 if day_of_week(year, month, 1) == day: total += 1
 return total

start = time.time()

total = months_start_range(0,1901,2000)

elapsed = time.time() - start

print("%s found in %s seconds") % (total,elapsed)
</code></pre>

This might you solve the problem.

It took around 0.068 seconds to solve it.

blocks|key|4494211|text|import+pandas+as+pd
from+datetime+import+date


start+=+date(1901,+1,+1)
end+=+date(2000,+12,+31)
d+=+pd.date_range(start,+end,+freq='MS').strftime('%25A')
s+=+pd.Series(d)

print(s.value_counts())|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|4494212|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>import pandas as pd
from datetime import date


start = date(1901, 1, 1)
end = date(2000, 12, 31)
d = pd.date_range(start, end, freq='MS').strftime('%A')
s = pd.Series(d)

print(s.value_counts())
</code></pre>

blocks|key|4493138|text|因此，我不是从日期的角度来处理这个问题，而是从倒计时的角度来处理这个问题。这是我的解决方案：。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4493139|days_1st+=+list()
day_counter+=+1
for+year+in+range(1900,+2001):
++++for+month+in+range(1,13):
++++++++#Skip+for+year+1900+as+count+starts+from+1901,+but+this+still+
++++++++#adds+the+days+hence+keeping+the+cycle+in+sync!
++++++++if+year+!=+1900:+
++++++++++++days_1st.append(day_counter)
++++++++if+month+==+4+or+month+==+6+or+month+==+9+or+month+==+11:
++++++++++++day_counter%2B=30
++++++++elif+month+==+2+and+((year+%25+100+==+0+and+year+%25+400+==+0)+or+(year+%25+100+!=+0+and+year+%25+4+==+0)):
++++++++++++day_counter%2B=29
++++++++elif+month+==+2:
++++++++++++day_counter%2B=28
++++++++else:
++++++++++++day_counter%2B=31
#+mod+7+because+since+the+day+the+counting+started+(1+Jan+1900+-+
#+Monday)+Every+7th+day+is+a+sunday!
days_sunday+=+list(filter(lambda+x:+x+%25+7+==+0,+days_1st))
print(len(days_sunday))+|code-block|syntax|javascript|4493140|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|$]]

So I approached this problem from not a date perspective but of a counting days.
Here's my solution:. 

<pre><code>days_1st = list()
day_counter = 1
for year in range(1900, 2001):
 for month in range(1,13):
 #Skip for year 1900 as count starts from 1901, but this still 
 #adds the days hence keeping the cycle in sync!
 if year != 1900: 
 days_1st.append(day_counter)
 if month == 4 or month == 6 or month == 9 or month == 11:
 day_counter+=30
 elif month == 2 and ((year % 100 == 0 and year % 400 == 0) or (year % 100 != 0 and year % 4 == 0)):
 day_counter+=29
 elif month == 2:
 day_counter+=28
 else:
 day_counter+=31
# mod 7 because since the day the counting started (1 Jan 1900 - 
# Monday) Every 7th day is a sunday!
days_sunday = list(filter(lambda x: x % 7 == 0, days_1st))
print(len(days_sunday)) 
</code></pre>

blocks|key|4493181|text|这里有一个不同的方法来解决这个问题|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4493182|public+static+void+main(String[]+args)+{

++++int+k+=+0;
++++//+String+months[]+=+{+"January",+"February",+"March",+"April",+"May",+"June",
++++//+"July",+"August",+"September",
++++//+"October",+"November",+"December"+};
++++String+Days[]+=+{+"Monday",+"Tuesday",+"Wednesday",+"Thursday",+"Friday",+"Saturday",+"Sunday"+};
++++int+MonthsDdays[]+=+{+31,+28,+31,+30,+31,+30,+31,+31,+30,+31,+30,+31+};
++++int+counter+=+0;
++++for+(int+t+=+1900;+t+<=+2000;+t%2B%2B)+{
++++++++MonthsDdays[1]=28;
++++++++if+(t+%25+4+==+0)+{
++++++++++++if+(t+%25+100+==+0)

++++++++++++{
++++++++++++++++if+(t+%25+400+==+0)
++++++++++++++++++++MonthsDdays[1]+=+29;
++++++++++++}+else+if+(t+%25+100+!=+0)
++++++++++++++++MonthsDdays[1]+=+29;

++++++++}

++++++++int+p+=+0;
++++++++while+(p+<+12)+{

++++++++++++for+(int+j+=+0;+j+<+MonthsDdays[p];+k%2B%2B,+j%2B%2B)+{
++++++++++++++++if+(k+==+7)
++++++++++++++++++++k+=+0;
++++++++++++++++if+(Days[k].equalsIgnoreCase("Sunday")+&&+j+==+0+&&+t+>+1900)+{
++++++++++++++++++++counter%2B%2B;

++++++++++++++++}
++++++++++++}
++++++++++++p%2B%2B;
++++++++}
++++}
++++System.out.println(counter);
}|code-block|syntax|javascript|4493183|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|$]]

Here is a different approach to tackle this question 

<pre><code>public static void main(String[] args) {

 int k = 0;
 // String months[] = { "January", "February", "March", "April", "May", "June",
 // "July", "August", "September",
 // "October", "November", "December" };
 String Days[] = { "Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday", "Sunday" };
 int MonthsDdays[] = { 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 };
 int counter = 0;
 for (int t = 1900; t &lt;= 2000; t++) {
 MonthsDdays[1]=28;
 if (t % 4 == 0) {
 if (t % 100 == 0)

 {
 if (t % 400 == 0)
 MonthsDdays[1] = 29;
 } else if (t % 100 != 0)
 MonthsDdays[1] = 29;

 }

 int p = 0;
 while (p &lt; 12) {

 for (int j = 0; j &lt; MonthsDdays[p]; k++, j++) {
 if (k == 7)
 k = 0;
 if (Days[k].equalsIgnoreCase("Sunday") &amp;&amp; j == 0 &amp;&amp; t &gt; 1900) {
 counter++;

 }
 }
 p++;
 }
 }
 System.out.println(counter);
}
</code></pre>

blocks|key|4494394|text|我尝试了数学方法，尽管我们可以使用日历函数。我首先计算月份的数学，以确定使用其他月份的月份的第一个日期之间的关系。此外，为了计算闰年，我计算了从3月到2月的年份。如果您想计算1901年的1月和2月，您可以编写一个单独的条件，并对删除2001年的1月和2月执行相同的操作。但是，在这种情况下，它们实际上并不重要，因为它们不是星期天，所以您可以删除此特定情况下的最后一个if条件。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4494395|++++#+Zero+is+Sunday+and+the+rest+of+the+days+are+according+to+mod7
++++#+li+stores+the+first+days+of+the+months+in+the+year+in+every+iteration
++++#+The+year+in+initial+li+is+1900+but+the+contents+are+Mar-1900+to+Feb-1901
++++#+At+the+end,+we+can+check+if+Jan+or+Feb+of+2001+contain+a+Sunday+and+remove+if+it+does
++++li,+cnt+=+[4,0,2,5,0,3,6,1,4,6,2,5],+0+
++++#+Could+also+initialize+li+from+by+the+same+method+as+below,+but+I+had+already+calculated+those
++++#+cnt+adds+number+of+zeros+in+every+iteration+(the+number+of+Sundays+in+every+year)+to+its+value
++++#+As+we+don't+count+for+the+year+1900+cnt=0,+else+initialize+cnt=li.count(0)
++++for+year+in+range(1901,2001):
++++++++if+year%254==0:
++++++++++++li[0]=li[8]=(li[11]%2B1)%257++++#Set+March+and+November+to+%2B1+value+than+last+Feb
++++++++else:
++++++++++++li[0]=li[8]=li[11]++++++++++#Set+March+and+November+to+same+value+as+last+Feb
++++++++#+The+following+values+of+other+months+will+depend+solely+on+their+own+March+value
++++++++#+You+can+check+the+Math+if+you+want+to
++++++++li[3]=li[11]=(li[0]%2B1)%257;li[6]=li[9]=(li[0]%2B2)%257;li[1]=li[4]=(li[0]%2B3)%257;li[2]=li[10]=(li[0]-2)%257;li[5]=(li[0]-1)%257;li[7]=(li[0]-3)%257
++++++++cnt+=+cnt+%2B+li.count(0)
++++#+This+is+to+remove+the+extra+two+months+of+the+year+2001+if+they+bother+the+answer
++++if+li[10]+==+0+or+li[11]+==+0:
++++++++cnt+=+cnt-1
++++print(cnt)|code-block|syntax|javascript|4494396|这是我在StackOverflow上的第一个答案，我希望我写得很好。;-)|4494397|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

I tried the Mathematical approach although we could use the calendar functions. 
I first calculated the math of the months to determine the relationships between the first dates of the months using the other months. Also, for simplicity in calculating leap years, I calculated the year from March to Feb. If you want to calculate for the Jan and Feb of 1901, you can write a separate condition, and do the same to remove Jan and Feb of 2001. However, in this case, they do not really matter as they are not Sundays, so you could remove the last if condition for this specific case.

<pre><code> # Zero is Sunday and the rest of the days are according to mod7
 # li stores the first days of the months in the year in every iteration
 # The year in initial li is 1900 but the contents are Mar-1900 to Feb-1901
 # At the end, we can check if Jan or Feb of 2001 contain a Sunday and remove if it does
 li, cnt = [4,0,2,5,0,3,6,1,4,6,2,5], 0 
 # Could also initialize li from by the same method as below, but I had already calculated those
 # cnt adds number of zeros in every iteration (the number of Sundays in every year) to its value
 # As we don't count for the year 1900 cnt=0, else initialize cnt=li.count(0)
 for year in range(1901,2001):
 if year%4==0:
 li[0]=li[8]=(li[11]+1)%7 #Set March and November to +1 value than last Feb
 else:
 li[0]=li[8]=li[11] #Set March and November to same value as last Feb
 # The following values of other months will depend solely on their own March value
 # You can check the Math if you want to
 li[3]=li[11]=(li[0]+1)%7;li[6]=li[9]=(li[0]+2)%7;li[1]=li[4]=(li[0]+3)%7;li[2]=li[10]=(li[0]-2)%7;li[5]=(li[0]-1)%7;li[7]=(li[0]-3)%7
 cnt = cnt + li.count(0)
 # This is to remove the extra two months of the year 2001 if they bother the answer
 if li[10] == 0 or li[11] == 0:
 cnt = cnt-1
 print(cnt)
</code></pre>

This was my first answer on StackOverflow, I hope I wrote well. ;-)

blocks|key|4494406|text|A+=+[31,28,31,30,31,30,31,31,30,31,30,31]
sunday+=0
gK+=+1
for+y+in+range(1901,2001):
++++if(y+%254+==0):
++++++++A[1]+=+29
++++else:
++++++++A[1]+=+28
++++for+m+in+range(len(A)):
++++++++for+d+in+range(1,A[m]%2B1):
++++++++++++if(gK+==6):
++++++++++++++++if(d==1):
++++++++++++++++++++sunday+%2B=1

++++++++++++++++gK+=0
++++++++++++else:
++++++++++++++++gK+=gK%2B1
print(sunday)|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|4494407|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>A = [31,28,31,30,31,30,31,31,30,31,30,31]
sunday =0
gK = 1
for y in range(1901,2001):
 if(y %4 ==0):
 A[1] = 29
 else:
 A[1] = 28
 for m in range(len(A)):
 for d in range(1,A[m]+1):
 if(gK ==6):
 if(d==1):
 sunday +=1

 gK =0
 else:
 gK =gK+1
print(sunday)
</code></pre>

blocks|key|4493353|text|我想我找到答案了。不过我不确定..你的逻辑是对的。但需要一点改进。我们需要首先计算星期二的数量，因为我们清楚地知道这是1900年1月1日的星期一。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4493354|months+=+{+"January":+31,
++++++++"February"+:+28,
++++++++"March"+:+31,
++++++++"April"+:+30,
++++++++"May"+:+31,
++++++++"June"+:+30,
++++++++"July"+:+31,
++++++++"August"+:+31,
++++++++"September"+:+30,
++++++++"October"+:+31,
++++++++"November"+:+30,
++++++++"December"+:+31}


for+month+in+months:
++++print(months[month])


tuesday_count+=+0
day+=+0+
extra_days+=+0
for+year+in+range(1901,+2001):
++++days_in_the_year+=+0
++++for+month+in+months:
++++++++day+%2B=+months[month]
++++++++days_in_the_year+%2B=+months[month]

++++++++if(+year+%25+4+==+0+and+month+==+'February'):
++++++++++++++++if+(year+%25+100+!=+0):
++++++++++++++++++++extra_days+%2B=+1
++++++++++++++++++++days_in_the_year+%2B=+1
++++++++++++++++++++day+%2B=+1
++++++++++++++++elif(year+%25+100+==0+and+year+%25+400+==0):
++++++++++++++++++++extra_days+%2B=+1
++++++++++++++++++++days_in_the_year+%2B=+1
++++++++++++++++++++day+%2B=+1
++++++++if(+(day)+%25+7++==+0):
++++++++++++++++tuesday_count+%2B=+1
++++print('No.+of+days+in+the+year',year,'are',days_in_the_year)

print('No.+of+Tuesdays+counted+so+far+is++=',+tuesday_count)
print('The+number+of+extra_days+because+of+the+leap+years+are:',extra_days)
#+print('extra_days+%25+7+=',+'25+%25+7+=',+extra_days+%25+7)
print('So,+there+were',+extra_days+//+7,+'extra_no_of_weeks+left+that+we+haven\'t+considered.+After+that,+it\'s+followed+by+--wed,+thu,+fri+and+sat+(we+don\'t+need+to+consider+that).\n+So,+the+total+number+of+Tuesdays+are',+tuesday_count%2B3+)

tuesday_count+%2B=+3

print('This+means+only+2+Sundays+that+have+followed')
sunday_count+=+tuesday_count+-+1
print('Since,+1901+Jan1+would+be+a+Tuesday,+we+need+to+subract+one+from+the+total+number+of+Sundays\n+So,+the+total+number+of+sundays+are:',+)
sunday_count=sunday_count-1
print(sunday_count)|code-block|syntax|javascript|4493355|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 think I got the answer. I am not sure though.. your logic was right. But needed a little improvement. We need to start off by counting the number of Tuesdays first as we clearly know that it was Monday on Jan 1, 1900.

<pre><code>
months = { "January": 31,
 "February" : 28,
 "March" : 31,
 "April" : 30,
 "May" : 31,
 "June" : 30,
 "July" : 31,
 "August" : 31,
 "September" : 30,
 "October" : 31,
 "November" : 30,
 "December" : 31}


for month in months:
 print(months[month])


tuesday_count = 0
day = 0 
extra_days = 0
for year in range(1901, 2001):
 days_in_the_year = 0
 for month in months:
 day += months[month]
 days_in_the_year += months[month]

 if( year % 4 == 0 and month == 'February'):
 if (year % 100 != 0):
 extra_days += 1
 days_in_the_year += 1
 day += 1
 elif(year % 100 ==0 and year % 400 ==0):
 extra_days += 1
 days_in_the_year += 1
 day += 1
 if( (day) % 7 == 0):
 tuesday_count += 1
 print('No. of days in the year',year,'are',days_in_the_year)

print('No. of Tuesdays counted so far is =', tuesday_count)
print('The number of extra_days because of the leap years are:',extra_days)
# print('extra_days % 7 =', '25 % 7 =', extra_days % 7)
print('So, there were', extra_days // 7, 'extra_no_of_weeks left that we haven\'t considered. After that, it\'s followed by --wed, thu, fri and sat (we don\'t need to consider that).\n So, the total number of Tuesdays are', tuesday_count+3 )

tuesday_count += 3

print('This means only 2 Sundays that have followed')
sunday_count = tuesday_count - 1
print('Since, 1901 Jan1 would be a Tuesday, we need to subract one from the total number of Sundays\n So, the total number of sundays are:', )
sunday_count=sunday_count-1
print(sunday_count)

</code></pre>

blocks|key|4493369|text|months=[31,28,31,30,31,30,31,31,30,31,30,31]
leap=[31,29,31,30,31,30,31,31,30,31,30,31]

sundays=0
start=2
for+y+in+range(25):
++for+nonleap+in+range+(3):
++++for+j+in+months:
++++++start=(start%2Bj)%257
++++++if+start+==+0:
++++++++sundays%2B=1
++++
++for+m+in+leap:
++++start=(start%2Bm)%257
++++if+start+==+0:
++++++sundays%2B=1
print+sundays|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|4493370|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>months=[31,28,31,30,31,30,31,31,30,31,30,31]
leap=[31,29,31,30,31,30,31,31,30,31,30,31]

sundays=0
start=2
for y in range(25):
 for nonleap in range (3):
 for j in months:
 start=(start+j)%7
 if start == 0:
 sundays+=1
 
 for m in leap:
 start=(start+m)%7
 if start == 0:
 sundays+=1
print sundays
</code></pre>

blocks|key|4494517|text|您已将day变量初始化为1，但1901年1月1日是星期二。我犯了同样的错误;-)|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4494518|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

you have initialized the day variable to 1 but the 1st Jan 1901 is a Tuesday. I made the same error ;-)

I was solving <a href="https://projecteuler.net/problem=19" rel="nofollow noreferrer">Project Euler #19</a>:

How many Sundays fell on the first of the month during the twentieth century (1 Jan 1901 to 31 Dec 2000)?

And here is the code :

<pre><code>months = { "January": 31,
 "February" : 28,
 "March" : 31,
 "April" : 30,
 "May" : 31,
 "June" : 30,
 "July" : 31,
 "August" : 31,
 "September" : 30,
 "October" : 31,
 "November" : 30,
 "December" : 31}

def countingSundays():
 day = 1
 sunday_count = 0

 for year in xrange(1901,2001):

 for m in months:

 day += months[m]
 if year % 4 == 0 and m == "February":
 day += 1
 if day % 7 == 0:
 sunday_count += 1

print "Sundays:", sunday_count
</code></pre>

The output of the program is 172 which is incorrect.
I searched the answer to be 171. 
So I wanted to know why am I getting the extra 1 Sunday ?

Project Euler #19, Python

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我在解决问题在二十世纪(1901年1月1日至2000年12月31日)每月的第一个星期天有多少个？下面是代码：months = { "January": 31,        "February" : 28,        "March" : 31,        "April" : 30,        "May" :...

问项目Euler #19，Python
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问项目Euler #19，Python
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