Project Euler解题汇总 041 ~ 050

问题41: 解答见按字典顺序生成所有的排列，此处不再重复。


问题42:How many triangle words does the list of common English words contain?
答　　案:162

import java.util.Scanner  
import java.io.File  
import scala.Math.sqrt

object Euler042 extends Application {
    var scan = new Scanner(new File("words.txt")).useDelimiter("\"(,\")?")
    var res = 0
    while(scan.hasNext()){
        var word = scan.next()
        var vle = word.foldLeft(0){ _+_-'A'+1 }
        var idx = sqrt(2*vle)
        if(idx*(idx+1) == 2*vle) res += 1
    }
    scan.close()
    println(res)
}

问题43:Find the sum of all pandigital numbers with an unusual sub-string divisibility property.
题目简介:数字1406357289具有以下性质：
　　1.它是0~9的一个排列
　　2.记di为它的第i为数字(从左至右,i从1开始),则:
　　　　　* d2d3d4=406 is divisible by 2
　　　　　* d3d4d5=063 is divisible by 3
　　　　　* d4d5d6=635 is divisible by 5
　　　　　* d5d6d7=357 is divisible by 7
　　　　　* d6d7d8=572 is divisible by 11
　　　　　* d7d8d9=728 is divisible by 13
　　　　　* d8d9d10=289 is divisible by 17
　　求所有具有上述性质的数字之和。
答　　案:16695334890

import eastsun.math.Util._

object Euler043 extends Application {
    val ps =Array(2,3,5,7,11,13,17)
    def suit(arr:Array[Int]):Boolean = {
        var num = arr.slice(0,3).foldLeft(0){ _*10+_ }
        for(idx <- 3 until arr.length){
            num =num%100*10 + arr(idx)
            if(num % ps(idx-3) != 0) return false
        }
        true
    }
    var arr = Array(0,1,2,3,4,5,6,7,8,9)
    var sum = 0L
    do{
        if(suit(arr)) sum += arr.foldLeft(0L){ _*10+_ }
    }while(nextPermutation(arr))
    println(sum)
}

问题44:Find the smallest pair of pentagonal numbers whose sum and difference is pentagonal.
题目简介:由公式Pn=n(3n1)/2生成的数称为五角数，前10个五角数依次为：
　　　　1, 5, 12, 22, 35, 51, 70, 92, 117, 145, ...
　　易见，P4 + P7 = 22 + 70 = 92 = P8。但他们的差P7-P4=70-22 =48不是一个五角数。
　　现在要求两个五角数Pj与Pk使得他们的和与差(绝对值)都是五角数，并且使得他们差的绝对值D = |Pj-Pk| 最小。输出D的值。
答　　案:5482660

object Euler044 extends Application {
    def isPenta(n:Int):Boolean = {
        val v = 6*n
        val s = Math.sqrt(v)
        s%3 == 2 && s*(s+1) == v
    }
    def penta(n:Int) = n*(3*n-1)/2
    
    var res = 0
    var k = 1
    while(res == 0){
        val pk = penta(k)
        var j = 1
        while(j < k && res == 0){
            val pj = penta(j)
            if(isPenta(pk+pj) && isPenta(pk-pj)) res = pk - pj
            j += 1
        }
        k += 1
    }
    println(res)
}

问题45:After 40755, what is the next triangle number that is also pentagonal and hexagonal?
题目简介:三角数，五角数，六角数的生成公式如下：
　　Triangle　　　Tn=n(n+1)/2　　　1, 3, 6, 10, 15, ...
　　Pentagonal　　Pn=n(3n-1)/2　　　1, 5, 12, 22, 35, ...
　　Hexagonal　　 Hn=n(2n-1)　　　　1, 6, 15, 28, 45, ...
　　可以验证，T285 = P165 = H143 = 40755
　　求下一个既是三角数又是五角数与六角数的数字。
题目分析:易见，六角数一定是三角数，因此只需寻找下一个是五角数的六角数即可。
答　　案:1533776805

//Scala
    Stream.from(144).map{ n => n*(2*(n:Long)-1) }.find{ v =>
        var sq = Math.sqrt(6*v).toLong
        (sq+1)%3 == 0 && sq*(sq+1) == 6*v
    }.get

问题46:What is the smallest odd composite that cannot be written as the sum of a prime and twice a square?
题目简介:Christian Goldbach提出过一个猜想：
　　任意一个奇合数都能写成一个平方数的两倍与一个素数之和。如：
　　　9 = 7 + 2×1^2
　　　15 = 7 + 2×2^2
　　　21 = 3 + 2×3^2
　　　25 = 7 + 2×3^2
　　　27 = 19 + 2×2^2
　　　33 = 31 + 2×1^2
　　但后来发现这个猜想是错误的。现在要求不满足上面猜想的最小的那个奇合数。
答　　案:5777

object Euler046 extends Application {
    var ps:Stream[Int] = Stream.cons(2,
                             Stream.from(3).filter{ n =>
                                 ps.takeWhile(p => p*p <= n).forall(n%_ !=0)
                             })
    def isPrime(n:Int) = ps.takeWhile{ p => p*p<=n }.forall{ n%_ !=0 }
    var res = 3
    var end = false
    do{
        do{
            res += 2
        }while(isPrime(res))
        val sq = Math.sqrt((res-2)/2)
        end = 1.to(sq).forall{ n => !isPrime(res-2*n*n) }
    }while(!end)
    println(res)
}

问题47:Find the first four consecutive integers to have four distinct primes factors.
题目简介:对于自然数n，记f(n)为满足下列条件的最小的那个数：f(n),f(n)+1,...,f(n)+n-1这连续n个数，每个数恰好有n个不同的素因子。
　　比如：
　　　　f(2) = 14,　　
　　　　　　14 = 2 ×7
　　　　　　15 = 3 ×5
　　　　f(3) = 644
　　　　　　644 = 2² ×7× 23
　　　　　　645 = 3× 5× 43
　　　　　　646 = 2× 17× 19
　　现在要求f(4)
答　　案:134043

import java.util.ArrayList

object Euler047 extends Application {
    val ps:Stream[Int] = Stream.cons(2,
                             Stream.from(3).filter{ n =>
                                 ps.takeWhile(p => p*p <= n).forall(n%_ !=0)
                             })
    def f(size:Int):Int = {
        var buf = new ArrayList[Int]
        buf.add(0)
        buf.add(0)
        var num = 2
        var cnt = 0
        while(cnt < size){
            var copy = num
            var first = ps.find{ copy%_ == 0 }.get
            while(copy % first == 0) copy = copy/first
            var vau = buf.get(copy)+1
            buf.add(vau)
            if(vau == size) cnt += 1
            else         cnt = 0
            num += 1
        }
        num - size
    }
    println(f(4))
}

问题48:Find the last ten digits of 1^1 + 2^2 + ... + 1000^1000.
题目简介:求1^1 + 2^2 + ... + 1000^1000末尾10位数字。
答　　案:9110846700

//Scala
val MAX = 10000000000L
def pow(n:Int):Long = 1.to(n).foldLeft(1L){ (p,i) => p*n%MAX }
1.to(1000).foldLeft(0L){ (s,i) => (s+pow(i))%MAX }

问题49:
Find arithmetic sequences, made of prime terms, whose four digits are permutations of each other.
题目简介:数1487, 4817, 8147有以下性质：
　　1.它们都是素数
　　2.这三个数成等差数列
　　3.除了位置不同，组成它们的数字一样。
求出四位数中满足上述条件的另三个数，按从小到大连着输出。
答　　案:296962999629

import java.util.Arrays.{ binarySearch => find}
object Euler049 extends Application {
    val ps:Stream[Int] = Stream.cons(2,
                             Stream.from(3).filter{ n =>
                                 ps.takeWhile(p => p*p <= n).forall(n%_ !=0)
                             })
    val pa = ps.takeWhile{ _<10000 }.dropWhile{ _<=1487 }.toArray
    
    def suit(a:Int,b:Int):Boolean = {
        var buf =new Array[Int](10)
        var tmp = a
        while(tmp != 0){
            buf(tmp%10) += 1
            tmp = tmp/10
        }
        tmp = b
        while(tmp != 0){
            buf(tmp%10) -= 1
            tmp = tmp/10
        }
        buf.forall{ _==0 }
    }
    var i = 0
    var res = ""
    while(res == "" && i < pa.length){
        var j = i+2
        while(res == "" && j < pa.length){
            var mid = (pa(i)+pa(j))/2
            if(find(pa,i,j,mid) >= 0 &&
               suit(pa(i),pa(j)) &&
               suit(pa(i),mid) ) res = ""+pa(i)+pa(j)
            j += 1
        }
        i += 1
    }
    println(res)
}

问题50:
Which prime, below one-million, can be written as the sum of the most consecutive primes?
题目简介:素数41能写成连续6个素数之和：
　　　　41 = 2 + 3 + 5 + 7 + 11 + 13
　　现在要求1,000,000以内的素数，能表示为最多的连续素数之和的那个数。
题目分析:我这儿采取的是类似“迭代加深搜索”的算法：首先对“连续素数”的个数做一个上界估计，记为len。然后看1000000以内有没有len个连续素数其和也为素数，并且在1000000以内。如果有，这这个和就是解；否则，将len减1，继续上述操作。
答　　案:997651

import java.util.Arrays.{binarySearch => indexOf}
import java.util.BitSet
import scala.collection.jcl.LinkedList

object Euler050 extends Application{
    //var start = System.currentTimeMillis
    /**
       get all primes below bound
       Sieve of Eratosthenes 
    */
    def getPrimes(bound:Int):Array[Int] = {
        if(bound <= 2) return Array()
        var set = new BitSet(bound)
        var idx = 2
        var lst = new LinkedList[Int]
        lst.add(2)
        while(idx < bound){
            var sp = lst.last
            var st = sp*2
            while(st <= bound){
                set.set(st)
                st += sp
            }
            st = sp+1
            while(st < bound && set.get(st)) st += 1
            if(st < bound) lst.add(st)
            idx = st
        }
        lst.toArray
    }
    
    val primes = getPrimes(1000000)
    val last = primes(primes.length-1)
    
    var len = primes.length/100
    var res = 0
    
    do{
        var start = 0
        var sum = start.until(start+len).foldLeft(0L){(s,i) => s+primes(i) }
        var idx = indexOf(primes,start+len,primes.length,sum.toInt)
        if(idx >= 0) res  = sum.toInt
        
        while(start+len < primes.length && sum <= last && res ==0){
            sum = sum +primes(start+len)-primes(start)
            idx = indexOf(primes,start+len,primes.length,sum.toInt)
            if(idx >= 0) res = sum.toInt
            start += 1
        }
        len -= 1
    }while(res == 0)
    
    //var end = System.currentTimeMillis
    
    println(res)
    //println("Use time(ms): "+(end-start))
}