Project Euler 16~20

Problem 16:

215 = 32768 and the sum of its digits is 3 + 2 + 7 + 6 + 8 = 26.

What is the sum of the digits of the number 21000?

a,ans=pow(2,1000),0
while a>0:
ans+=a%10
a//=10
print(ans)
Answer:1366

Problem 17:

If the numbers 1 to 5 are written out in words: one, two, three, four, five, then there are 3 + 3 + 5 + 4 + 4 = 19 letters used in total.

If all the numbers from 1 to 1000 (one thousand) inclusive were written out in words, how many letters would be used?

NOTE: Do not count spaces or hyphens. For example, 342 (three hundred and forty-two) contains 23 letters and 115 (one hundred and fifteen) contains 20 letters. The use of "and" when writing out numbers is in compliance with
British usage.

a=[0 for i in range(1000+10)]
a[1]=3
a[2]=3
a[3]=5
a[4]=4
a[5]=4
a[6]=3
a[7]=5
a[8]=5
a[9]=4
a[10]=3
a[11]=6
a[12]=6
a[13]=8
a[14]=8
a[15]=7
a[16]=7
a[17]=9
a[18]=8
a[19]=8
a[20]=6
a[30]=6
a[40]=5
a[50]=5
a[60]=5
a[70]=7
a[80]=6
a[90]=6
a[100]=7
a[1000]=8
ans=0
for i in range(1,20+1):
ans+=a[i]
for i in range(21,100):
a[i]=a[i//10*10]+a[i%10]
ans+=a[i]
for i in range(100,1000):
if i%100==0:
ans+=a[i//100]+a[100]
else:
ans+=a[i//100]+a[100]+3+a[i%100]
ans+=a[1]+a[1000]
print(ans)

Answer:21124

Problem 18:

By starting at the top of the triangle below and moving to adjacent numbers on the row below, the maximum total from top to bottom is 23.

3
7 4

2 4 6

8 5 9 3
That is, 3 + 7 + 4 + 9 = 23.

Find the maximum total from top to bottom of the triangle below:

75

95 64

17 47 82

18 35 87 10

20 04 82 47 65

19 01 23 75 03 34

88 02 77 73 07 63 67

99 65 04 28 06 16 70 92

41 41 26 56 83 40 80 70 33

41 48 72 33 47 32 37 16 94 29

53 71 44 65 25 43 91 52 97 51 14

70 11 33 28 77 73 17 78 39 68 17 57

91 71 52 38 17 14 91 43 58 50 27 29 48

63 66 04 68 89 53 67 30 73 16 69 87 40 31

04 62 98 27 23 09 70 98 73 93 38 53 60 04 23
NOTE: As there are only 16384 routes, it is possible to solve this problem by trying every route. However,

Problem 67, is the same challenge with a triangle containing one-hundred rows; it cannot be solved by brute force, and requires a clever method! ;o)

a=[[75],

[95,64],

[17,47,82],

[18,35,87,10],

[20,4,82,47,65],

[19,1,23,75,3,34],

[88,2,77,73,7,63,67],

[99,65,4,28,6,16,70,92],

[41,41,26,56,83,40,80,70,33],

[41,48,72,33,47,32,37,16,94,29],

[53,71,44,65,25,43,91,52,97,51,14],

[70,11,33,28,77,73,17,78,39,68,17,57],

[91,71,52,38,17,14,91,43,58,50,27,29,48],

[63,66,4,68,89,53,67,30,73,16,69,87,40,31],

[4,62,98,27,23,9,70,98,73,93,38,53,60,4,23]]

def dfs(x,y,ans):
if (x==14):
return ans
return max([dfs(x+1,y,ans+a[x+1][y]),dfs(x+1,y+1,ans+a[x+1][y+1])])
print(dfs(0,0,75))

Answer:1074

Problem 19:

You are given the following information, but you may prefer to do some research for yourself.

1 Jan 1900 was a Monday.
Thirty days has September,

April, June and November.

All the rest have thirty-one,

Saving February alone,

Which has twenty-eight, rain or shine.

And on leap years, twenty-nine.
A leap year occurs on any year evenly divisible by 4, but not on a century unless it is divisible by 400.
How many Sundays fell on the first of the month during the twentieth century (1 Jan 1901 to 31 Dec 2000)?

year,month,day,week,ans,f=1900,1,1,1,0,0
d=[0,31,28,31,30,31,30,31,31,30,31,30,31]
def days(x,y):
if x==2:
if y%400==0 or y%4==0 and y%100!=0:
return 29
else:
return 28
else:
return d[x]
while 1:
if year==2000 and month==12 and day==31:
break
day+=1
week+=1
if week==8:
week=1
if day>days(month,year):
day=1
month+=1
if month>12:
month=1
year+=1
if year==1901 and month==1 and day==1:
f=1
if f and day==1 and week==7:
ans+=1
print(ans)

Answer:171

Problem 20:

n! means n

(n


1)


...


3


2


1

For example, 10! = 10

9


...


3


2


1 = 3628800,

and the sum of the digits in the number 10! is 3 + 6 + 2 + 8 + 8 + 0 + 0 = 27.

Find the sum of the digits in the number 100!

a,ans=1,0
for i in range(1,100+1):
a*=i
while a>0:
ans+=a%10
a//=10
print(ans)

Answer:648

By Charlie Pan
Apr 16,2014