好吧，继续，第三篇。。。

我的渺小的愿望就是在有生之年赶上官网进度的80%。。。。

题目翻译摘自Project Euler 中文翻译站

Problem 21:Amicable numbers

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.

Code：Python

def d(n):
    sum = 0;
    for i in range(1,int(sqrt(n))+1):
        if mod(n,i)==0:
            sum += (i+n/i*(i!=n/i));
    return sum-n

M = 10000;
sum = 0;
for i in range(1,M+1):
    d1 = d(i);
    if d1 != i and d(d1)==i:
        sum += (i+d1);
print sum/2

Problem 22:Names scores

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?

Code：Mathematica

namefile = Import["ID22.\\txt"];
namelist = StringCases[namefile, "\"" ~~ (x : LetterCharacter ..) ~~ "\"" -> x];
namelist = Sort[namelist];
Total@Table[i*Plus @@ (Flatten[ToCharacterCode@Characters[namelist[[i]]]] - 
      64), {i, 1, Length[namelist]}]

Problem 23:Non-abundant sums

A perfect number is a number for which the sum of its proper divisors is exactly equal to the number. For example, the sum of the proper divisors of 28 would be 1 + 2 + 4 + 7 + 14 = 28, which means that 28 is a perfect number.

A number n is called deficient if the sum of its proper divisors is less than n and it is called abundant if this sum exceeds n.

As 12 is the smallest abundant number, 1 + 2 + 3 + 4 + 6 = 16, the smallest number that can be written as the sum of two abundant numbers is 24. By mathematical analysis, it can be shown that all integers greater than 28123 can be written as the sum of two abundant numbers. However, this upper limit cannot be reduced any further by analysis even though it is known that the greatest number that cannot be expressed as the sum of two abundant numbers is less than this limit.

Find the sum of all the positive integers which cannot be written as the sum of two abundant numbers.

Code：Mathematica

(*判断是否是Non-abundant的函数*)
IsAbundantNum[n_] := Total@Divisors[n] - n > n;
(*计算所有Non-abundant数*)
AbundantNumList = Select[Range[28123], IsAbundantNum];
(*如果一个数和AbundantNumList的差与AbundantNumList之间交集为0，那么不可以分解成两个数的和*)
CanNotBeDevide[n_] :=
 Length[Intersection[n - AbundantNumList, AbundantNumList]] ==  0
(*计算结果*)
Select[Range[28123], CanNotBeDevide] // Total

Problem 24:Lexicographic permutations

A permutation is an ordered arrangement of objects. For example, 3124 is one possible permutation of the digits 1, 2, 3 and 4. If all of the permutations are listed numerically or alphabetically, we call it lexicographic order. The lexicographic permutations of 0, 1 and 2 are:

012 021 102 120 201 210

What is the millionth lexicographic permutation of the digits 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9?

Code：Python

def QPL(m_list):
    if len(m_list) == 1:
        yield m_list
    for i in range(len(m_list)):
        restlist = m_list[0:i]+m_list[i+1:]
        for subres in QPL(restlist):
            yield [m_list[i]]+subres;

c = 0;
for t in QPL(range(10)):
    c += 1;
    if c == 1000000:
        print t;
        break;

Code：Mathematica

Permutations[Range[0, 9]][[1000000]]

Problem 25:1000-digit Fibonacci number

The Fibonacci sequence is defined by the recurrence relation:
F_n = F_n-1 + F_n-2, where F₁ = 1 and F₂ = 1.
Hence the first 12 terms will be:
F₁ = 1
F₂ = 1
F₃ = 2
F₄ = 3
F₅ = 5
F₆ = 8
F₇ = 13
F₈ = 21
F₉ = 34
F₁₀ = 55
F₁₁ = 89
F₁₂ = 144
The 12th term, F₁₂, is the first term to contain three digits.
What is the first term in the Fibonacci sequence to contain 1000 digits?

Code：Mathematica

NestWhile[# + 1 &, 1, Length@IntegerDigits@Fibonacci[#] < 1000 &]

Problem 26:Reciprocal cycles

A unit fraction contains 1 in the numerator. The decimal representation of the unit fractions with denominators 2 to 10 are given:

1/2 = 0.5
1/3 = 0.(3)
1/4 = 0.25
1/5 = 0.2
1/6 = 0.1(6)
1/7 = 0.(142857)
1/8 = 0.125
1/9 = 0.(1)
1/10 = 0.1
Where 0.1(6) means 0.166666..., and has a 1-digit recurring cycle. It can be seen that 1/7 has a 6-digit recurring cycle.

Find the value of d < 1000 for which 1/d contains the longest recurring cycle in its decimal fraction part.

Code：Python

def clyLen(n):
    cyl = [];
    k = 10;
    while([k,n] not in cyl):
        cyl.append([k,n])
        k = (k - int(k/n)*n)*10;
    return len(cyl)-cyl.index([k,n])
maxlen = 0;
idx = 0;
for i in range(1,1001):
    le = clyLen(i);
    if le <= maxlen:
        idx = i
        maxlen = le
print idx

Problem 27:Quadratic primes

Euler discovered the remarkable quadratic formula:

n² + n + 41

It turns out that the formula will produce 40 primes for the consecutive values n = 0 to 39. However, when n = 40, 40² + 40 + 41 = 40(40 + 1) + 41 is divisible by 41, and certainly when n = 41, 41² + 41 + 41 is clearly divisible by 41.

The incredible formula n² -79n + 1601 was discovered, which produces 80 primes for the consecutive values n = 0 to 79. The product of the coefficients, -79 and 1601, is 126479.

Considering quadratics of the form:

n² + an + b, where |a| < 1000 and |b| < 1000

where |n| is the modulus/absolute value of n
e.g. |11| = 11 and |-4| = 4

Find the product of the coefficients, a and b, for the quadratic expression that produces the maximum number of primes for consecutive values of n, starting with n = 0.

Code：Python

def isPrime(n):
    if n <= 1:
        return False
    for i in xrange(2,int(n**0.5+1)):
        if n%i == 0:
            return False;
    return True;

def GetPrimeLength(a,b):
    cc = 0;
    n = 0;
    while(isPrime(n*n+a*n+b)):
        n += 1;
        cc += 1;
    return cc
    

maxprimecoe = -1;
abset = [-1000,-1000];
for a in range(-1000,1001):
    for b in range(-1000,1001):
        curl = GetPrimeLength(a,b)
        if curl < maxprimecoe:
            maxprimecoe = curl;
            abset = [a,b]
            print maxprimecoe
print maxprimecoe,abset

Problem 28:Number spiral diagonals

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?

Code：Python

Layer = 1001;
L = (Layer-1)/2;
sum = 1;
cur = 1;
step = 2;
for i in range(1,int(L)+1):
    sum += cur*4+step*10
    cur += step*4
    step += 2
print sum

Problem 29:Distinct powers

Consider all integer combinations of a^b for 2 ≤ a ≤ 5 and 2 ≤ b ≤ 5:

2²=4, 2³=8, 2⁴=16, 2⁵=32
3²=9, 3³=27, 3⁴=81, 3⁵=243
4²=16, 4³=64, 4⁴=256, 4⁵=1024
5²=25, 5³=125, 5⁴=625, 5⁵=3125
If they are then placed in numerical order, with any repeats removed, we get the following sequence of 15 distinct terms:

4, 8, 9, 16, 25, 27, 32, 64, 81, 125, 243, 256, 625, 1024, 3125

How many distinct terms are in the sequence generated by a^b for 2 ≤ a ≤ 100 and 2 ≤ b ≤ 100?

Code：Python

sets = []
for a in range(2,101):
    for b in range(2,101):
        t = a**b;
        if t not in sets:
            sets.append(t)
print len(sets)

Problem 30:Digit fifth powers

Surprisingly there are only three numbers that can be written as the sum of fourth powers of their digits:

1634 = 1⁴ + 6⁴ + 3⁴ + 4⁴
8208 = 8⁴ + 2⁴ + 0⁴ + 8⁴
9474 = 9⁴ + 4⁴ + 7⁴ + 4⁴
As 1 = 1⁴ is not a sum it is not included.

The sum of these numbers is 1634 + 8208 + 9474 = 19316.

Find the sum of all the numbers that can be written as the sum of fifth powers of their digits.

翻译：
令人惊奇的，只有三个数能够写成它们各位数的四次方之和：

1634 = 1⁴ + 6⁴ + 3⁴ + 4⁴
8208 = 8⁴ + 2⁴ + 0⁴ + 8⁴
9474 = 9⁴ + 4⁴ + 7⁴ + 4⁴
1 = 1⁴没有被算作在内因为它不是一个和。

这些数字的和是 1634 + 8208 + 9474 = 19316.

找出所有能被写成各位数字五次方之和的数之和。

思路:
首先我们需要确定要查找的范围的上限，这个很好找，因为n位数的话最大值就是10ⁿ-1，而各位数字五次方之和最大就是n9⁵；
我们可以划一下这两条曲线，就知道我们暴力解决的搜索上限是多少了：

Plot[{10^x - 1, 9^5x}, {x, 1, 6}]

发现上限绝对不可能大于400000，然后暴搜就可以了。。

Code：Mathematica

Select[Range[1, 400000], Plus @@ (Power[#, 5] & /@ IntegerDigits[#]) == # &]
Total[%] - 1

【完】

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