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Euler problems/171 to 180

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Solution:
 
Solution:
{{sect-stub}}
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<haskell>
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#include <stdio.h>
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static int result = 0;
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#define digits 20
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static long long fact[digits+1];
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static const long long precision = 1000000000;
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static const long long precision_mult = 111111111;
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#define maxsquare 64 /* must be a power of 2 > digits * 9^2 */
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static inline int issquare( int n )
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{
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for( int step = maxsquare/2, i = step;;)
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{
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if( i*i == n ) return i;
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if( !( step >>= 1 ) ) return -1;
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if( i*i > n ) i -= step;
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else i += step;
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}
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}
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static inline void dodigit( int d, int nr, int sum, long long c, int s )
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{
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if( d )
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for( int n = 0; n <= nr; c *= ++n, s += d, sum += d*d )
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dodigit( d-1, nr - n, sum, c, s );
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else if( issquare( sum ) > 0 )
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result = ( s * ( fact[digits] / ( c * fact[nr] ) )
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/ digits % precision * precision_mult
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+ result ) % precision;
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}
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int main( void )
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{
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fact[0] = 1;
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for( int i = 1; i < digits+1; i++ ) fact[i] = fact[i-1]*i;
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dodigit( 9, digits, 0, 1, 0 );
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printf( "%d\n", result );
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return 0;
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}
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problem_171 = main
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</haskell>
   
 
== [http://projecteuler.net/index.php?section=problems&id=172 Problem 172] ==
 
== [http://projecteuler.net/index.php?section=problems&id=172 Problem 172] ==

Revision as of 13:49, 24 February 2008

Contents

1 Problem 171

Finding numbers for which the sum of the squares of the digits is a square.

Solution:

#include <stdio.h>
 
static int result = 0;
 
#define digits 20
static long long fact[digits+1];
static const long long precision      = 1000000000;
static const long long precision_mult =  111111111;
 
#define maxsquare 64 /* must be a power of 2 > digits * 9^2 */
 
static inline int issquare( int n )
{
    for( int step = maxsquare/2, i = step;;)
    {
        if( i*i == n )        return i;
        if( !( step >>= 1 ) ) return -1;
        if( i*i > n )         i -= step;
        else                  i += step;
    }
}
 
static inline void dodigit( int d, int nr, int sum, long long c, int s )
{
    if( d )
        for( int n = 0; n <= nr; c *= ++n, s += d, sum += d*d )
            dodigit( d-1, nr - n, sum, c, s );
    else if( issquare( sum ) > 0 )
        result = ( s * ( fact[digits] / ( c * fact[nr] ) )
                 / digits % precision * precision_mult
                 + result ) % precision;
}
 
int main( void )
{
    fact[0] = 1;
    for( int i = 1; i < digits+1; i++ ) fact[i] = fact[i-1]*i;
    dodigit( 9, digits, 0, 1, 0 );
    printf( "%d\n", result );
    return 0;
}
problem_171 = main

2 Problem 172

Investigating numbers with few repeated digits.

Solution:

factorial n = product [1..toInteger n]
 
fallingFactorial x n = product [x - fromInteger i | i <- [0..toInteger n - 1] ]
 
choose n k = fallingFactorial n k `div` factorial k
 
-- how many numbers can we get having d digits and p positions
p172 0 _ = 0
p172 d p 
    | p < 4 = d^p
    | otherwise = 
        (p172' p) +  p*(p172' (p-1)) + (choose p 2)*(p172' (p-2)) + (choose p 3)*(p172' (p-3))
    where
    p172' = p172 (d-1)
 
problem_172= (p172 10 18) * 9 `div` 10

3 Problem 173

Using up to one million tiles how many different "hollow" square laminae can be formed? Solution:

problem_173=
    let c=div (10^6) 4
        xm=floor$sqrt $fromIntegral c
        k=[div c x|x<-[1..xm]]
    in  sum k-(div (xm*(xm+1)) 2)

4 Problem 174

Counting the number of "hollow" square laminae that can form one, two, three, ... distinct arrangements.

Solution: This was my C++ code, published here without my permission nor any attribution, shame on whoever put it here. henk263

5 Problem 175

Fractions involving the number of different ways a number can be expressed as a sum of powers of 2. Solution:

sternTree x 0=[]
sternTree x y=
    m:sternTree y n  
    where
    (m,n)=divMod x y
findRat x y
    |odd l=take (l-1) k++[last k-1,1]
    |otherwise=k
    where
    k=sternTree x y
    l=length k
p175 x y= 
    init$foldl (++) "" [a++","|
    a<-map show $reverse $filter (/=0)$findRat x y]
problems_175=p175 123456789 987654321
test=p175 13 17

6 Problem 176

Rectangular triangles that share a cathetus. Solution:

--k=47547 
--2*k+1=95095 = 5*7*11*13*19
lst=[5,7,11,13,19]
primes=[2,3,5,7,11]
problem_176 =
    product[a^b|(a,b)<-zip primes (reverse n)]
    where
    la=div (last lst+1) 2
    m=map (\x->div x 2)$init lst
    n=m++[la]

7 Problem 177

Integer angled Quadrilaterals.

Solution: This C++ solution is stolen from balakrishnan. Check out the forum if you want to see his solution to the problem

8 Problem 178

Step Numbers

Solution: This C++ solution is stolen from balakrishnan. Check out the forum if you want to see his solution to the problem

9 Problem 179

Consecutive positive divisors.

10 Problem 180

Rational zeros of a function of three variables. Solution:

import Data.Ratio
 
{-
  After some algebra, we find:
   f1 n x y z = x^(n+1) + y^(n+1) - z^(n+1)
   f2 n x y z = (x*y + y*z + z*x) * ( x^(n-1) + y^(n-1) - z^(n-1) )
   f3 n x y z = x*y*z*( x^(n-2) + y^(n-2) - z^(n-2) )
   f n x y z = f1 n x y z + f2 n x y z - f3 n x y z
   f n x y z = (x+y+z) * (x^n+y^n-z^n)
Now the hard part comes in realizing that n can be negative.  
Thanks to Fermat, we only need examine the cases n = [-2, -1, 1, 2]
Which leads to:
 
f(-2) z = xy/sqrt(x^2 + y^2)
f(-1) z = xy/(x+y)
f(1)  z = x+y
f(2)  z = sqrt(x^2 + y^2)
 
-}
 
unique ::  Eq(a) => [a] -> [a]
unique [] = []
unique (x:xs) | elem x xs = unique xs
              | otherwise = x : unique xs
 
-- Not quite correct, but I don't care about the zeros
ratSqrt :: Rational -> Rational
ratSqrt x = 
  let a = floor $ sqrt $ fromIntegral $ numerator x
      b = floor $ sqrt $ fromIntegral $ denominator x
      c = (a%b) * (a%b)
  in if x == c then (a%b) else 0
 
-- Not quite correct, but I don't care about the zeros
reciprocal :: Rational -> Rational
reciprocal x 
  | x == 0 = 0
  | otherwise = denominator x % numerator x
 
problem_180 =
  let order = 35
      range :: [Rational]            
      range = unique [ (a%b) | b <- [1..order], a <- [1..(b-1)] ]
      fm2,fm1,f1,f2 :: [[Rational]]
      fm2 = [[x,y,z] | x<-range, y<-range, 
            let z = x*y * reciprocal (ratSqrt(x*x+y*y)), elem z range]
      fm1 = [[x,y,z] | x<-range, y<-range, 
            let z = x*y * reciprocal (x+y), elem z range]
      f1  = [[x,y,z] | x<-range, y<-range, 
            let z = (x+y), elem z range]
      f2  = [[x,y,z] | x<-range, y<-range, 
            let z = ratSqrt(x*x+y*y), elem z range]            
      result = sum $ unique $ map (\x -> sum x) (fm2++fm1++f1++f2)
  in (numerator result + denominator result)