Starting with the number 1 and moving to the right in a clockwise direction a 5 by 5 spiral is formed as follows:

21 22 23 24 25
20  7  8  9 10
19  6  1  2 11
18  5  4  3 12
17 16 15 14 13

It can be verified that the sum of the numbers on the diagonals is 101.

What is the sum of the numbers on the diagonals in a 1001 by 1001 spiral formed in the same way?

观察数字1,3,5,7,9,13,17,21,25,易知对角线数字递增规律为连续4个差2(3 + 2 = 5),连续4个差4(9 + 4 =13)…连续4个差(n-1);

当然我们也可以从中得出递推关系,观察右上对角线数字,1, 9, 25为1, 3, 5的平方,所有我们很容易知道f(n)的最右上的数字为 \(n^2\),则f(n)为f(n-2)加上最外围四个数字:
f(n) = f(n - 2) + 4 \(n^2\) - 6(n - 1);
一个简单的等差数列,最后求得:
\begin{eqnarray*} f(n) = \frac{2}{3}(n - 1)^{3} + \frac{5}{2}(n - 1)^2 + \frac{13}{3}(n - 1) + 1 \end{eqnarray*}

669171001

Source:C++

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)?

很直接的,查下表1901/1/1是星期二,此时设表示周几的变量numOfDay = 2,那么1901/2/1是周几？只要numOfDay加上一月份的天数31,再模7得5知为周五.依此法检测1901~2000每月第一天是周几。注意:别忘了2月份的天数需要是否闰年的判断.

171

Source:C++

The series, \(1^1\) + \(2^2\) + \(3^3\) + … + \(10^{10}\) = 10405071317.

Find the last ten digits of the series, \(1^1\) + \(2^2\) + \(3^3\) + … + \(1000^{1000}\).

对于直接大数据操作的,直接运算即可,不支持的只要对结果用模取最后10位进行操作即可.

9110846700

Source:C++

Using names.txt (right click and ‘Save Link/Target As…’), a 46K text file containing over five-thousand first names, begin by sorting it into alphabetical order. Then working out the alphabetical value for each name, multiply this value by its alphabetical position in the list to obtain a name score.

For example, when the list is sorted into alphabetical order, COLIN, which is worth 3 + 15 + 12 + 9 + 14 = 53, is the 938th name in the list. So, COLIN would obtain a score of 938 53 = 49714.

What is the total of all the name scores in the file?

没什么好说的,先读入数据,然后排序,最后进行计算,效率关键在于排序.

871198282

Source:C++

Let d(n) be defined as the sum of proper divisors of n (numbers less than n which divide evenly into n). If d(a) = b and d(b) = a, where a b, then a and b are an amicable pair and each of a and b are called amicable numbers.

For example, the proper divisors of 220 are 1, 2, 4, 5, 10, 11, 20, 22, 44, 55 and 110; therefore d(220) = 284. The proper divisors of 284 are 1, 2, 4, 71 and 142; so d(284) = 220.

Evaluate the sum of all the amicable numbers under 10000.

按题意暴力求解,但我们还是能在细节进行一些优化.

1. 我们只寻找amicable pair numbers中小的数,但相加时把两者都加上.
2. getDivisorSum的优化
   1. 直观的,循环只要到sqrt(n)就行,因为除数是对应的,n = a * b,由反证法易知一数小于等于sqrt(n)，另一数大于等于sqrt(n);
   2. 我们考虑求质数,公式( \ref{ref1} ),公式( \ref{ref2} )明显成立,公式( \ref{ref3} )左右两边展开相等可证
      

      getDivisorSum(p1*p2*..p_n) = 1+ p1 + p2 + ... pn + p1p2 + .. + p(n-1)p(n) + ... + p1p2..pn 
                                 = (1 + p1)(1 + p2)...(1 + pn) 
                                 = getDivisorSum(p1) * getDivisorSum(p2) * ... * getDivisorSum(pn)
      

\begin{eqnarray} getDivisorSum(p) = (p + 1)\label{ref1}\ \end{eqnarray} \begin{eqnarray} getDivisorSum(p^i) = 1 + p^2 + … + p^i = \frac{p^{i+1}-1}{p-1}\label{ref2}\ \end{eqnarray} \begin{eqnarray} getDivisorSum(p1*p2*..p_n) = getDivisorSum(p1) * getDivisorSum(p2) * …\label{ref3} \end{eqnarray}

31626

Source:C++

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)

   3                3             3          23
  7 4             7  4         20   19  
 2 4 6    ->    10 13 15   ->           ->  
8 5 9 3       

1074

Source:C++迭代 C++递归

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.

我们知道<1000的都可由,one,two…ninteen,twenty,thirty…ninety,hundred,and 这些英文字母表达出来，并将one~nineteen的字母数保存在数组 a[19]={3, 3, 5, 4, 4, 3, 5, 5, 4, 3, 6, 6, 8, 8, 7, 7, 9, 8, 8} 中，twenty,thirty..ninety的字母数保存在数组 b[8]＝{6, 6, 5, 5, 5, 7, 6, 6} 中，简单分类计算sum：

1. \(i\in[1,20)\)
   sum = sum+ a[i - 1]
2. \(i\in[20,100)\)
   1. sum = sum + b[i / 10 - 2] \(i\in\) {20,30,..90}
   2. sum = sum + b[i / 10 - 2] + a[i % 10 - 1] \(i\in\) {21,22,…99}
3. \(i\in[100,1000)\)
   1. sum = sum + a[i / 100 - 1] + len(“hundred”) \(i\in\) {100,200,300..900}
   2. sum = sum + a[i / 100 - 1] + len(“hundred”) + len(“and”) + a[i % 100 - 1]; \(i\in\) {101-119,201-219,…901-919}
   3. sum = sum + a[i / 100 - 1] + len(“hundred”) + len(“and”) + b[i / 10 % 10 - 2]; \(i\in\) {120,130,140..990}
   4. sum = sum + a[i / 100 - 1] + len(“hundred”) + len(“and”) + a[i % 10 - 1] + b[i / 10 % 10 - 2]; \(i\in\) {121,122…999}
4. \(i = 1000\)
   sum = sum + len(“one”) + len(“thousand”);

21124

Source:C++

Starting in the top left corner of a 2*2 grid, and only being able to move to the right and down, there are exactly 6 routes to the bottom right corner.

How many such routes are there through a 20*20 grid?

设左上角坐标为(0,0),右下角坐标为(n,n),则有

递归式:

path(i, j) = path(i-1, j) + path(i, j-1)

因为只能向右或向下运动，故处于某点(i,j)时的路径数为与此点相邻的(i-1,j)和(i,j-1)路径数之和.

递归终止条件为当点位于最右(最下)时，其路径数是唯一的,只能一直向下（向右),即path(n,i) = 1, path(i,n) = 1

从数学的角度看,只需要用到高中的排列组合.从源到目的地共需要移动n + n步,求其路径数即求哪些步骤向右哪些步骤向下，即：

path = \(C_{2n}^{n}\)

当然对于a*b的格子路径数也是同样的做法

137846528820

Source:C++

The Fibonacci sequence is defined by the recurrence relation:

\(F_{n}\) = \(F_{n-1}\) + \(F_{n-2}\) , where \(F_1\) = 1 and \(F_2\) = 1. Hence the first 12 terms will be:

\(F_1\) = 1
\(F_2\) = 1
\(F_3\) = 2
\(F_4\) = 3
\(F_5\) = 5
\(F_6\) = 8
\(F_7\) = 13
\(F_8\) = 21
\(F_9\) = 34
\(F_{10}\) = 55
\(F_{11}\) = 89
\(F_{12}\) = 144
The 12th term, \(F_{12}\) , is the first term to contain three digits.

What is the first term in the Fibonacci sequence to contain 1000 digits?

没什么好说的,大数据处理

4782

Source:C++