前日在网上闲逛，发现了这个有意思的网站 Project Euler。这个网站给出了一系列数学相关的题目，你可以使用编程去解答。
引用
What is Project Euler?
Project Euler is a series of challenging mathematical/computer programming problems that will require more than just mathematical insights to solve. Although mathematics will help you arrive at elegant and efficient methods, the use of a computer and programming skills will be required to solve most problems.
The motivation for starting Project Euler, and its continuation, is to provide a platform for the inquiring mind to delve into unfamiliar areas and learn new concepts in a fun and recreational context.
Project Euler is a series of challenging mathematical/computer programming problems that will require more than just mathematical insights to solve. Although mathematics will help you arrive at elegant and efficient methods, the use of a computer and programming skills will be required to solve most problems.
The motivation for starting Project Euler, and its continuation, is to provide a platform for the inquiring mind to delve into unfamiliar areas and learn new concepts in a fun and recreational context.
蛮有趣的，我准备在空余时间依次做做这些题。这系列帖子用于列出我所做出的题（每个贴包括5~15道问题的答案）：对于简单的题，直接给出程序代码及答案；对于我认为比较复杂的题目，只在该贴给出答案，然后另开新帖给出详细点的解答。
对于解题所用的编程语言，那些简单的题可能只给出 Scala代码（ 可以直接使用Scala控制台运行，最后一行的运行结果即为答案）；复杂一些的题可能会使用Java/C/Mathmatica/Matlab（注：Mathmatica与Matlab是两个常用的数学工具）。
对于所写的代码，在能够解决问题的前提下，尽量做到简单&漂亮， 所以所给的代码效率上未必是最佳的，能够解决问题足矣。
问题1:
Add all the natural numbers below one thousand that are multiples of 3 or 5.
求小于1000的自然数中3或5的倍数之和。
答案:233168
Stream.range(0,1000).filter{ n => n%3==0||n%5==0 }.foldLeft(0){ _+_ }
(0 until 1000).filter{ n => n%3==0||n%5==0 }.sum
问题2:
Find the sum of all the even-valued terms in the Fibonacci sequence which do not exceed four million.
对于数列f(n):
f(1) =1,f(2) =2
f(n) =f(n-1)+f(n-2) (n>2)
求数列f(n)中不大于4000,000且为偶数项数之和。
答案:4613732
var fib:Stream[Int] = Stream.cons(1,Stream.cons(2,
fib.zip(fib.tail).map{case(x,y) => x+y}
))
fib.takeWhile{_<=4000000}.filter{_%2==0}.foldLeft(0){_+_}
Stream.iterate((1,2)){ case(x,y) => (y,x+y) }.map{ _._1 }
.takeWhile{ _ <= 4000000 }.filter{ _%2 == 0 }.sum
问题3: Find the largest prime factor of a composite number.
求600851475143的最大质因数。
答案:6857
var num = 600851475143L
var res = 2
while(res != num) if(num%res == 0) num = num/res else res += 1
res
Stream.iterate((600851475143L,2)){ case(n,p) => if(n%p == 0) (n/p,p) else (n,p+1) }
.takeWhile{ case(n,p) => n>=p }.last._1
问题4:
Find the largest palindrome made from the product of two 3-digit numbers.
求能分解为两个三位数乘积的最大回文数。
答案:906609
def isPalin(n:Int):Boolean = {
var inv =0
var tmp =n
while(tmp != 0){
inv = inv*10 + tmp%10
tmp = tmp/10
}
inv == n
}
var res =0
for(i <- 100 to 999;j <- 100 to 999;if isPalin(i*j))
if(i*j > res) res = i*j
res
(for(x <- 100 to 999;y <- 100 to 999;s = "" + x*y;if s == s.reverse) yield x*y).max
问题5: What is the smallest number divisible by each of the numbers 1 to 20?
求1,2,..,19,20的最小公倍数。
答案:232792560
def gcd(a:Long,b:Long):Long = if(a==0) b else gcd(b%a,a)
1.to(20).foldLeft(1L){ (a,b) => a*b/gcd(a,b)}
问题6: Find the difference between the sum of the squares of the first one hundred natural numbers and the square of the sum.
求1至100和的平方与1至100的平方和之差。
答案:25164150
var sumOfSquare = 1.to(100).foldLeft(0){ (a,b) => a+b*b }
var sum = 1.to(100).foldLeft(0){ _+_ }
var squareOfSum = sum*sum
squareOfSum-sumOfSquare
(for(x <- 1 to 100;y <- 1 to 100;if x != y) yield x*y).sum
问题7: Find the 10001st prime.
求第10001个素数（质数）。
答案:104743
var ps:Stream[Int] = Stream.cons(2,
Stream.from(3).filter{ n =>
ps.takeWhile(p => p*p <= n).forall(n%_ !=0)
})
ps(10000)
问题8:
Discover the largest product of five consecutive digits in the 1000-digit number.
在给定的1000个数中，求出连续5个数的最大乘积。
答案:40824
var nums ="""73167176531330624919225119674426574742355349194934
96983520312774506326239578318016984801869478851843
85861560789112949495459501737958331952853208805511
12540698747158523863050715693290963295227443043557
66896648950445244523161731856403098711121722383113
62229893423380308135336276614282806444486645238749
30358907296290491560440772390713810515859307960866
70172427121883998797908792274921901699720888093776
65727333001053367881220235421809751254540594752243
52584907711670556013604839586446706324415722155397
53697817977846174064955149290862569321978468622482
83972241375657056057490261407972968652414535100474
82166370484403199890008895243450658541227588666881
16427171479924442928230863465674813919123162824586
17866458359124566529476545682848912883142607690042
24219022671055626321111109370544217506941658960408
07198403850962455444362981230987879927244284909188
84580156166097919133875499200524063689912560717606
05886116467109405077541002256983155200055935729725
71636269561882670428252483600823257530420752963450""".replaceAll("\\s+","").map{ _.asDigit }
var max =0
for(i <- 0 until nums.size-5){
var value =nums.slice(i,i+5).foldLeft(1){ (a,b) => a*b }
if(value>max) max =value
}
max
(for(i <- 0 until nums.size-5) yield nums.slice(i,i+5).product).max
问题9:
Find the only Pythagorean triplet, {a, b, c}, for which a + b + c = 1000.
求满足a+b+c = 1000的勾股数的积。（注：勾股数指满足a^2+b^2=c^2的自然数a,b,c）
答案:31875000
for{a <- 1 until 500
b <- a until 500
if a*a + b*b == (1000-a-b)*(1000-a-b)
} yield a*b*(1000-a-b)
问题10:
Calculate the sum of all the primes below two million.
求2,000,000以内所有素数的和。
答案:142913828922
//这里用的求素数算法同题7一样
//但这个问题数据量稍大了点，该代码需要运行15秒左右
//如果遇到更大的数据，我会采用稍复杂一点的算法，会快很多（快一到两个数量级）
var ps:Stream[Int] = Stream.cons(2,
Stream.from(3).filter(n =>
ps.takeWhile(p => p*p <= n).forall(n%_ !=0)
)
)
ps.takeWhile(_ < 2000000).foldLeft(0L){ _+_ }
问题11: What is the greatest product of four numbers on the same straight line in the 20 by 20 grid?
对于给定的20×20的数字方阵，求相邻4个数之积的最大值。（相邻数包括同一行、同一列或同一对角线上的连续相邻4个数）
答案:70600674
var src ="""08 02 22 97 38 15 00 40 00 75 04 05 07 78 52 12 50 77 91 08
49 49 99 40 17 81 18 57 60 87 17 40 98 43 69 48 04 56 62 00
81 49 31 73 55 79 14 29 93 71 40 67 53 88 30 03 49 13 36 65
52 70 95 23 04 60 11 42 69 24 68 56 01 32 56 71 37 02 36 91
22 31 16 71 51 67 63 89 41 92 36 54 22 40 40 28 66 33 13 80
24 47 32 60 99 03 45 02 44 75 33 53 78 36 84 20 35 17 12 50
32 98 81 28 64 23 67 10 26 38 40 67 59 54 70 66 18 38 64 70
67 26 20 68 02 62 12 20 95 63 94 39 63 08 40 91 66 49 94 21
24 55 58 05 66 73 99 26 97 17 78 78 96 83 14 88 34 89 63 72
21 36 23 09 75 00 76 44 20 45 35 14 00 61 33 97 34 31 33 95
78 17 53 28 22 75 31 67 15 94 03 80 04 62 16 14 09 53 56 92
16 39 05 42 96 35 31 47 55 58 88 24 00 17 54 24 36 29 85 57
86 56 00 48 35 71 89 07 05 44 44 37 44 60 21 58 51 54 17 58
19 80 81 68 05 94 47 69 28 73 92 13 86 52 17 77 04 89 55 40
04 52 08 83 97 35 99 16 07 97 57 32 16 26 26 79 33 27 98 66
88 36 68 87 57 62 20 72 03 46 33 67 46 55 12 32 63 93 53 69
04 42 16 73 38 25 39 11 24 94 72 18 08 46 29 32 40 62 76 36
20 69 36 41 72 30 23 88 34 62 99 69 82 67 59 85 74 04 36 16
20 73 35 29 78 31 90 01 74 31 49 71 48 86 81 16 23 57 05 54
01 70 54 71 83 51 54 69 16 92 33 48 61 43 52 01 89 19 67 48"""
def getValue(array:Array[Int],row:Int,col:Int):Int ={
var res =0
if(col+4 <= 20){
var pc = array.slice(row*20+col,row*20+col+4).foldLeft(1){ _*_ }
res =res max pc
}
if(row+4 <= 20){
var pr =1
for(n <- 0 until 4) pr = pr*array((row+n)*20+col)
res =res max pr
}
if(row+4<=20 && col+4<=20){
var pd1 =1
var pd2 =1
for(n <-0 until 4){
pd1 = pd1*array((row+n)*20+col+n)
pd2 = pd2*array((row+n)*20+col+3-n)
}
res =res max pd1 max pd2
}
res
}
var nums =src.split("\\s+").map{ _.foldLeft(0){ (a,b) => a*10+b-'0' } }
var res =0
for(row <- 0 until 20;col <- 0 until 20)
res =res max getValue(nums,row,col)
res
object Euler11 extends Application {
val src = """08 02 22 97 38 15 00 40 00 75 04 05 07 78 52 12 50 77 91 08
49 49 99 40 17 81 18 57 60 87 17 40 98 43 69 48 04 56 62 00
81 49 31 73 55 79 14 29 93 71 40 67 53 88 30 03 49 13 36 65
52 70 95 23 04 60 11 42 69 24 68 56 01 32 56 71 37 02 36 91
22 31 16 71 51 67 63 89 41 92 36 54 22 40 40 28 66 33 13 80
24 47 32 60 99 03 45 02 44 75 33 53 78 36 84 20 35 17 12 50
32 98 81 28 64 23 67 10 26 38 40 67 59 54 70 66 18 38 64 70
67 26 20 68 02 62 12 20 95 63 94 39 63 08 40 91 66 49 94 21
24 55 58 05 66 73 99 26 97 17 78 78 96 83 14 88 34 89 63 72
21 36 23 09 75 00 76 44 20 45 35 14 00 61 33 97 34 31 33 95
78 17 53 28 22 75 31 67 15 94 03 80 04 62 16 14 09 53 56 92
16 39 05 42 96 35 31 47 55 58 88 24 00 17 54 24 36 29 85 57
86 56 00 48 35 71 89 07 05 44 44 37 44 60 21 58 51 54 17 58
19 80 81 68 05 94 47 69 28 73 92 13 86 52 17 77 04 89 55 40
04 52 08 83 97 35 99 16 07 97 57 32 16 26 26 79 33 27 98 66
88 36 68 87 57 62 20 72 03 46 33 67 46 55 12 32 63 93 53 69
04 42 16 73 38 25 39 11 24 94 72 18 08 46 29 32 40 62 76 36
20 69 36 41 72 30 23 88 34 62 99 69 82 67 59 85 74 04 36 16
20 73 35 29 78 31 90 01 74 31 49 71 48 86 81 16 23 57 05 54
01 70 54 71 83 51 54 69 16 92 33 48 61 43 52 01 89 19 67 48"""
val nms = src.lines.toArray.map{ line => line.trim.split("\\s+").map{ _.toInt } }
def valueAt(row: Int,col: Int): Int =
if(row >= 0 && row < nms.size &&
col >= 0 && col < nms(row).size) nms(row)(col)
else 0
val ots = (0,1)::(1,0)::(1,1)::(1,-1)::Nil
val res = (for(row <- 0 until nms.size; col <- 0 until nms(row).size; (x,y) <- ots)
yield (for(s <- 0 until 4) yield valueAt(row+s*x,col+s*y)).product
).max
println(res)
}
问题12: What is the value of the first triangle number to have over five hundred divisors?
求第一个因子个数超过500的三角数。(三角数t(n) =n*(n+1)/2)
答案:76576500
var ts = Stream.cons(0,Stream.from(1).map{ n =>
var t =n*(n+1)/2
var f =0
var k =1
while(k*k