我正在解决Project Euler＃19：
在二十世纪的第一个月（1901年1月1日至2000年12月31日），有多少个星期日下降？
以下是代码：
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
程序的输出是172，这是不正确的。 我搜索的答案是171。 所以我想知道为什么我得到额外的1周日？
答案 0 :(得分：3)
你正在迭代months dict，期望它按照月份的顺序进行迭代，但是不会对dicts进行排序，因此你可以按错误的顺序获取月份。 / p>
由于您实际上并不需要月份名称，因此您只需将months列为月份长度即可。
答案 1 :(得分：1)
您应该使用datetime库，它将自动处理所有闰年信息：
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]
答案 2 :(得分：1)
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 > 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)
这可能会解决问题。
解决它需要大约0.068秒。
答案 3 :(得分：1)
这里是解决此问题的另一种方法
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++) {
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 < 12) {
for (int j = 0; j < MonthsDdays[p]; k++, j++) {
if (k == 7)
k = 0;
if (Days[k].equalsIgnoreCase("Sunday") && j == 0 && t > 1900) {
counter++;
}
}
p++;
}
}
System.out.println(counter);
}
答案 4 :(得分：0)
你犯的错误：
正确的计划是：
from collections import OrderedDict
months = OrderedDict( [("January",31),("February", 28),("March",31),
("April", 30), ("May", 31), ("June", 30),
("July", 31), ("August", 31), ("September", 30),
("October", 31), ("November", 30), ("December", 31)] )
days = ['Tuesday','Wednesday', 'Thursday','Friday','Saturday', 'Sunday', 'Monday']
day = 0
sunday_count = 0
def isLeap(year): #https://en.wikipedia.org/wiki/Leap_year#Algorithm
leap = True
if year % 4 != 0:
leap = False
elif year % 100 != 0:
leap = True
elif year % 400 != 0:
leap = False
return leap
for year in xrange(1901,2001):
leap = isLeap(year)
for m in months:
dayName = days[day%7]
if dayName == "Sunday":
sunday_count += 1
#print year, m, dayName
day += months[m]
if leap == True and m == "February":
day += 1
print sunday_count
# print 171
另外，有些日子：
1901 January Tuesday
1901 February Friday
1901 March Friday
1901 April Monday
1901 May Wednesday
1901 June Saturday
1901 July Monday
1901 August Thursday
1901 September Sunday
...
答案 5 :(得分：0)
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())
答案 6 :(得分：0)
所以我从日期的角度来看这个问题，而不是计算天数。 这是我的解决方案：
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))
答案 7 :(得分：0)
尽管我们可以使用日历功能，但我尝试了数学方法。 我首先计算了月份的数学运算，以确定使用其他月份的月份的第一天之间的关系。另外，为简化计算leap年，我计算了从3月到2月的年份。如果要计算1901年1月和2月，则可以编写一个单独的条件，并执行相同的操作以删除2001年1月和2月。但是，在这种情况下，它们并不重要，因为它们不是星期日，因此您可以删除此特定情况的最后一个if条件。
# 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)
这是我对StackOverflow的第一个答案，希望我写得不错。 ;-）
答案 8 :(得分：0)
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)
答案 9 :(得分：0)
==> Python解决方案
euler19.py
normal_year = [31,28,31,30,31,30,31,31,30,31,30,31]
leap_year = [31,29,31,30,31,30,31,31,30,31,30,31]
years = [ normal_year ] * 100
for i in range(3, len(years), 4) :
years[i] = leap_year
current_day = (0+365) % 7
sundays = 0
for y in years :
for m in y :
if current_day % 7 == 6:
sundays += 1
current_day += m%7
print (sundays)
答案 10 :(得分：0)
我想我得到了答案。不过，我不确定。您的逻辑是否正确。但是需要一点改进。首先，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，星期一。
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)
答案 11 :(得分：0)
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
答案 12 :(得分：0)
请注意，问题将1900年的第一天定义为星期一，而您将1901年的第一天定义为星期一。
months = [31,28,31,30,31,30,31,31,30,31,30,31]
def countingSundays():
day = 1
sunday_count = 0
for year in range(1900,1901):
for m in months:
day += m
if (year % 4 == 0 and m == 28):
day += 1
for year in range(1901,2001):
for m in months:
day += m
if (year % 4 == 0 and m == 28):
day += 1
if day % 7 == 0:
sunday_count += 1
return sunday_count
print ("Sundays:", countingSundays())
答案 13 :(得分：0)
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;-）