Project Euler解题汇总 013 ~ 022

　　[b][color=darkred]前言[/color][/b]:[url=http://projecteuler.net/]Project Euler[/url]是一个很有趣的网站，上面提供了一系列有一定挑战性的数学问题，这些问题如果单凭手工计算做不了几个，但如果使用编程并采用适当的算法就可以解决之。如果你对数学感兴趣并且会编程不妨去试试。
　　这篇帖子接着[url=http://eastsun.iteye.com/blog/203289]Euler Project解题汇总 001 ~ 012[/url]的内容继续将我最近做的一些题的答案贴上来。[color=red]注意[/color]，有一个不同的地方是：上一个贴中的[url=http://eastsun.iteye.com/category/30003]Scala[/url]代码都能够直接复制到[url=http://eastsun.iteye.com/category/30003]Scala[/url]控制台运行。由于某些原因，[color=red]以后贴出的Scala代码有些可能需要先保存为文本文件，然后编译运行，而不能直接复制到控制台来运行[/color]。这样的代码会包含一个完整的类[color=red]EulerXXX[/color](XXX表示题号)，并且最后一行的代码一定是一个[color=blue]println[/color]语句。

[url=http://projecteuler.net/index.php?section=problems&id=13]问题13[/url]:[color=orange]Find the first ten digits of the sum of one-hundred 50-digit numbers.[/color]
求给定的100个50位数之和的前10个数字。
[color=red]答案[/color]:5537376230

/**
  &#Euler013.scala
  @author: Eastsun
*/
object Euler013 extends Application {                 
    var src = """37107287533902102798797998220837590246510135740250
                 46376937677490009712648124896970078050417018260538
                 74324986199524741059474233309513058123726617309629
                 91942213363574161572522430563301811072406154908250
                 23067588207539346171171980310421047513778063246676
                 89261670696623633820136378418383684178734361726757
                 28112879812849979408065481931592621691275889832738
                 44274228917432520321923589422876796487670272189318
                 47451445736001306439091167216856844588711603153276
                 70386486105843025439939619828917593665686757934951
                 62176457141856560629502157223196586755079324193331
                 64906352462741904929101432445813822663347944758178
                 92575867718337217661963751590579239728245598838407
                 58203565325359399008402633568948830189458628227828
                 80181199384826282014278194139940567587151170094390
                 35398664372827112653829987240784473053190104293586
                 86515506006295864861532075273371959191420517255829
                 71693888707715466499115593487603532921714970056938
                 54370070576826684624621495650076471787294438377604
                 53282654108756828443191190634694037855217779295145
                 36123272525000296071075082563815656710885258350721
                 45876576172410976447339110607218265236877223636045
                 17423706905851860660448207621209813287860733969412
                 81142660418086830619328460811191061556940512689692
                 51934325451728388641918047049293215058642563049483
                 62467221648435076201727918039944693004732956340691
                 15732444386908125794514089057706229429197107928209
                 55037687525678773091862540744969844508330393682126
                 18336384825330154686196124348767681297534375946515
                 80386287592878490201521685554828717201219257766954
                 78182833757993103614740356856449095527097864797581
                 16726320100436897842553539920931837441497806860984
                 48403098129077791799088218795327364475675590848030
                 87086987551392711854517078544161852424320693150332
                 59959406895756536782107074926966537676326235447210
                 69793950679652694742597709739166693763042633987085
                 41052684708299085211399427365734116182760315001271
                 65378607361501080857009149939512557028198746004375
                 35829035317434717326932123578154982629742552737307
                 94953759765105305946966067683156574377167401875275
                 88902802571733229619176668713819931811048770190271
                 25267680276078003013678680992525463401061632866526
                 36270218540497705585629946580636237993140746255962
                 24074486908231174977792365466257246923322810917141
                 91430288197103288597806669760892938638285025333403
                 34413065578016127815921815005561868836468420090470
                 23053081172816430487623791969842487255036638784583
                 11487696932154902810424020138335124462181441773470
                 63783299490636259666498587618221225225512486764533
                 67720186971698544312419572409913959008952310058822
                 95548255300263520781532296796249481641953868218774
                 76085327132285723110424803456124867697064507995236
                 37774242535411291684276865538926205024910326572967
                 23701913275725675285653248258265463092207058596522
                 29798860272258331913126375147341994889534765745501
                 18495701454879288984856827726077713721403798879715
                 38298203783031473527721580348144513491373226651381
                 34829543829199918180278916522431027392251122869539
                 40957953066405232632538044100059654939159879593635
                 29746152185502371307642255121183693803580388584903
                 41698116222072977186158236678424689157993532961922
                 62467957194401269043877107275048102390895523597457
                 23189706772547915061505504953922979530901129967519
                 86188088225875314529584099251203829009407770775672
                 11306739708304724483816533873502340845647058077308
                 82959174767140363198008187129011875491310547126581
                 97623331044818386269515456334926366572897563400500
                 42846280183517070527831839425882145521227251250327
                 55121603546981200581762165212827652751691296897789
                 32238195734329339946437501907836945765883352399886
                 75506164965184775180738168837861091527357929701337
                 62177842752192623401942399639168044983993173312731
                 32924185707147349566916674687634660915035914677504
                 99518671430235219628894890102423325116913619626622
                 73267460800591547471830798392868535206946944540724
                 76841822524674417161514036427982273348055556214818
                 97142617910342598647204516893989422179826088076852
                 87783646182799346313767754307809363333018982642090
                 10848802521674670883215120185883543223812876952786
                 71329612474782464538636993009049310363619763878039
                 62184073572399794223406235393808339651327408011116
                 66627891981488087797941876876144230030984490851411
                 60661826293682836764744779239180335110989069790714
                 85786944089552990653640447425576083659976645795096
                 66024396409905389607120198219976047599490197230297
                 64913982680032973156037120041377903785566085089252
                 16730939319872750275468906903707539413042652315011
                 94809377245048795150954100921645863754710598436791
                 78639167021187492431995700641917969777599028300699
                 15368713711936614952811305876380278410754449733078
                 40789923115535562561142322423255033685442488917353
                 44889911501440648020369068063960672322193204149535
                 41503128880339536053299340368006977710650566631954
                 81234880673210146739058568557934581403627822703280
                 82616570773948327592232845941706525094512325230608
                 22918802058777319719839450180888072429661980811197
                 77158542502016545090413245809786882778948721859617
                 72107838435069186155435662884062257473692284509516
                 20849603980134001723930671666823555245252804609722
                 53503534226472524250874054075591789781264330331690"""
        var res = src.split("\\D+")
                     .map{ _.foldLeft(0:BigInt){ (a,b) => a*10+(b-'0') } }
                     .reduceLeft[BigInt]{ _+_ }
                     .toString
                     .substring(0,10)
        println(res)
}

[url=http://projecteuler.net/index.php?section=problems&id=14]问题14[/url]: [color=orange]Find the longest sequence using a starting number under one million. [/color]
对于正整数n，定义如下变换：
　　　n -> n/2　　如果n是偶数
　　　n -> 3*n+1　如果n是奇数
对13，反复使用上述变换，可以得到一个自然数序列：
　　　13 → 40 → 20 → 10 → 5 → 16 → 8 → 4 → 2 → 1
这个序列的长度为10。
对任意一个自然数，都能通过上述方式产生一个序列，该序列的结尾为1。
现在求1,000,000以内，产生序列最长的那个正整数。
[color=red]答案[/color]:837799

    //Scala
    def lenOfPath(n:Long):Int ={
        var len =1
        var copy =n
        while(copy != 1){
            copy = if(copy%2 == 0) copy/2 else copy*3+1
            len += 1
        }
        len
    }
    var res =0
    var len =0
    for(n <- 1 until 1000000){
        val tmp =lenOfPath(n)
        if(tmp > len){
            len =tmp
            res =n
        }
    }
    res

def lenOfPath(n: Long) = 
    Iterator.iterate(n){ k => if(k%2==0) k/2 else 3*k+1 }.takeWhile{ _!=1 }.size+1

(1 until 1000000).map{ n => (n,lenOfPath(n)) }.sortBy(_._2).last

[url=http://projecteuler.net/index.php?section=problems&id=15]问题15[/url]:[color=orange]Starting in the top left corner in a 20 by 20 grid, how many routes are there to the bottom right corner?[/color]
一个20×20的方格，从左上角走到右下角，每次走一格，并且方向只能向右或向下。问一共有多少种走法。
[color=red]答案[/color]:137846528820

    //Scala
    //Hint:就是求组合数C(40,20)
    1.to(20).foldLeft(1:BigInt){ (i,n) => i*(n+20)/n }

[url=http://projecteuler.net/index.php?section=problems&id=16]问题16[/url]:[color=orange] What is the sum of the digits of the number 2^1000?[/color]
求2的1000幂的各位数字之和。
[color=red]答案[/color]:1366

//Scala
var num = (1:BigInt)<<1000
num.toString.foldLeft(0){ _+_-'0' }

[url=http://projecteuler.net/index.php?section=problems&id=17]问题17[/url]:[color=orange]How many letters would be needed to write all the numbers in words from 1 to 1000?[/color]
从1到1000（包括）的英文表示共需要多少英文字母？
[color=olive]　　比如342用英语表述为：three hundred and forty-two
　　一共含23个字母（空格与连线符不算）[/color]
[color=red]答案[/color]:21124

    //Scala
    val tens = Array(0,0,6,6,5,5,5,7,6,6)
    val units = Array(0,3,3,5,4,4,3,5,5,4,
                      3,6,6,8,8,7,7,9,8,8)
    def lenOf(n:Int):Int ={
        val high =n/100
        val low =n%100
        var len =0
        if(high !=0) len = 7 + units(high)
        if(high!=0 && low!=0) len += 3
        if(low <20) len+units(low)
        else len+tens(low/10)+units(low%10)
    }
    1.to(999).foldLeft(11){ (a,b) => a+lenOf(b) }

[url=http://projecteuler.net/index.php?section=problems&id=18]问题18[/url]:[color=orange]Find the maximum sum travelling from the top of the triangle to the base.[/color]
给定一个数字组成的三角形，从三角形的顶点往下走一直到底部，将所经过的数字相加，求这个和的最大值。
[color=red]答案[/color]:1074

/**
   &#Euler018.scala
   @author Eastsun
*/
object Euler018 extends Application {
    var src ="""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
             """
    var num = src.split("(?m:$\\s*)")
                 .map{ _.split("\\s+")
                        .map{ _.foldLeft(0){ _*10+_-'0' } } 
                 }
    def path(row:Int,col:Int):Int =
        if(row ==num.length-1) num(row)(col)
        else num(row)(col) +path(row+1,col).max(path(row+1,col+1))

    println(path(0,0))
}

[url=http://projecteuler.net/index.php?section=problems&id=19]问题19[/url]:[color=orange]How many Sundays fell on the first of the month during the twentieth century?[/color]
已知1900年一月一日是星期一，求从1901.1.1到2000.12.31这段时间，既是星期天又是一个月的第一天的日子有多少天？
[color=red]答案[/color]:171

    //Scala
    def daysOfMonth(year:Int,month:Int):Int = month match{
        case 4|6|9|11        => 30
        case 1|3|5|7|8|10|12 => 31
        case _  => if(year%400==0||(year%4==0&&year%100!=0)) 29 else 28
    }
    var offset = 1.to(12).foldLeft(1){(s,m) => s+daysOfMonth(1900,m)}
    var res = 0
    for(year <- 1901 to 2000;month <- 1 to 12){
        if(offset%7 == 0) res += 1
        offset += daysOfMonth(year,month)
    }
    res

[url=http://projecteuler.net/index.php?section=forum&id=20]问题20[/url]:[color=orange]Find the sum of digits in 100![/color]
求100的阶乘的各位数字之和。
[color=red]答案[/color]:648

//Scala
1.to(100).foldLeft(1:BigInt){ _*_ }.toString.foldLeft(0){ _+_-'0' }

[url=http://projecteuler.net/index.php?section=problems&id=21]问题21[/url]:[color=orange]Evaluate the sum of all amicable pairs under 10000.[/color]
记d(n)为正整数n的真约数（小于其自身的因子）之和。
称正整数对（a，b）为亲和数，如果a，b满足：
　　　　a != b 且 d(a) =b, d(b) =a
求小于10000的所有亲和数之和。
[color=red]答案[/color]:31626

//Scala
    def sumOfDivisor(n:Int):Int ={
        if(n <= 1) return 0
        var sum =1
        var div =2
        while(div*div < n){
            if(n%div == 0) sum =sum+div+n/div
            div += 1
        }
        if(div*div == n) sum+div else sum
    }
    var res =0
    for(n <- 2 until 10000){
        val sd =sumOfDivisor(n)
        if(sd<n && sumOfDivisor(sd)==n) res =res+sd+n
    }
    res

[url=http://projecteuler.net/index.php?section=problems&id=22]问题22[/url]:[color=orange]What is the total of all the name scores in the file of first names?[/color]
给定一个文本文件[url=http://projecteuler.net/project/names.txt]names.txt[/url]，里面含有超过5000个的英文名字。你需要首先将它们按字典顺序排序，然后对于每一个名字给定一个权重，每个名字的得分为其权重与其在列表中的位置（这个位置从1算起）的乘积。计算所有名字得分之和。
比如名字“COLIN“，它的权重是3(3表示C是字母表中的第三个) + 15 + 12 + 9 + 14 = 53，它在列表中的位置是938th，故“COLIN“的得分是938 × 53 = 49714。
[color=red]答案[/color]:871198282

/**
   &#Euler022.scala
   @author Eastsun
*/
import java.util.Scanner
import java.io.File
import scala.collection.jcl.LinkedList
import scala.util.Sorting._

object Euler022 extends Application {
    var scan = new Scanner(new File("names.txt")).useDelimiter("\"(,\")?")
    var list = new LinkedList[String]
    while(scan.hasNext()) list.add(scan.next())
    scan.close()
    var array =list.toArray
    quickSort(array)
    var res = array.zipWithIndex.foldLeft(0){(v,s) =>
                  v +(s._2+1)*s._1.foldLeft(0){ _+_-'A'+1 }
              }
    println(res)
}