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),0while a>0:    ans+=a%10    a//=10print(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]=3a[2]=3a[3]=5a[4]=4a[5]=4a[6]=3a[7]=5a[8]=5a[9]=4a[10]=3a[11]=6a[12]=6a[13]=8a[14]=8a[15]=7a[16]=7a[17]=9a[18]=8a[19]=8a[20]=6a[30]=6a[40]=5a[50]=5a[60]=5a[70]=7a[80]=6a[90]=6a[100]=7a[1000]=8ans=0for 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=1900,1,1,1,0,0d=[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+=1print(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=1,0for i in range(1,100+1):    a*=iwhile a>0:    ans+=a%10    a//=10print(ans)

Answer:648

By Charlie Pan

Apr 16,2014