 1
I have 3 strings. I want a way to store for each string a collection of digits.
Or I want for each array to store the products of digits.
731671765313
306249192251
196744265747
I think of a Map<String, List<Integer>> products = new HashMap<>();
But I don't know how to write the code so for each string I must store a collection of digits.
Thanks in advance!
Best regards!
OCAJP 7
For example, given the three lines you show, what collection of digits do you want?
And as for "for each array to store the products of digits.", that is even more unclear.
How many products do you want for each string?
Do you want one (presumably the product of all the single digits in the line)?
Or perhaps the produce of the first and second, and the product of the second and third, etc.?
Or the product of the first 3, and the second three, etc.?
You can use a Stream to calculate the transverse product of digits.
(731671765313, [7, 3,2,1,6,7,1,7,6,5,3,1,3]
306249192251, [3,0,6,2,4,9,1,9,2,2,5,1]
196744265747, [1,9,6,7,4,4,2,6,5,7,4.7] )
or
(731671765313, 7*3*1*6*7*1*7*6*5*3*1*3
306249192251, 3*0*6*2*4*9*1*9*2*2*5*1
196744265747, 1*9*6*7*4*4*2*6*5*7*4*7 )
My solution is complicated for what must be done.
Output will be maximum of product.
OCAJP 7
https://coderanch.com/t/678032/java/Typevariableintdouble
The most difficult part of wild life photography is getting them to sign the model release.
That may mean the posts will be in the wrong order.
It's about problem 8 of Euler project:
Largest product in a series
Problem 8
Published on Friday, 11th January 2002, 08:00 pm; Solved by 255293; Difficulty rating: 5%
The four adjacent digits in the 1000digit number that have the greatest product are 9 × 9 × 8 × 9 = 5832.
73167176531330624919225119674426574742355349194934
96983520312774506326239578318016984801869478851843
85861560789112949495459501737958331952853208805511
12540698747158523863050715693290963295227443043557
66896648950445244523161731856403098711121722383113
62229893423380308135336276614282806444486645238749
30358907296290491560440772390713810515859307960866
70172427121883998797908792274921901699720888093776
65727333001053367881220235421809751254540594752243
52584907711670556013604839586446706324415722155397
53697817977846174064955149290862569321978468622482
83972241375657056057490261407972968652414535100474
82166370484403199890008895243450658541227588666881
16427171479924442928230863465674813919123162824586
17866458359124566529476545682848912883142607690042
24219022671055626321111109370544217506941658960408
07198403850962455444362981230987879927244284909188
84580156166097919133875499200524063689912560717606
05886116467109405077541002256983155200055935729725
71636269561882670428252483600823257530420752963450
Find the thirteen adjacent digits in the 1000digit number that have the greatest product. What is the value of this product?
Output is: 2.3514624E10
This is correct version.
Below will ne the wrong version: the only difference will be the type of variables.
My question is: how is avoid an such mistake?
Output is:2091059712
I think the output should be negative number, but it's not.
OCAJP 7
Rajith Pemabandu wrote:You need a long definitely to achieve a solution. (long (2)^63 to ((2^63)1) range) 64 bits.
Misplaced parens. Should be ((2^63))
Raising (2) to an odd power results in a negative number, but raising it to an even power would have resulted in a positive value.
((2^63)) will always be negative, even if you change 63 to 62.
The most difficult part of wild life photography is getting them to sign the model release.
Because the compiler is designed to look for breaches of the rules of syntax. It cannot be expected to look for logic errors. That is what programmers are there for.Dana Ucaed wrote:Why the compiler didn't say anything?
Your outputs follow the rules of integer arithmetic. You have suffered an overflow error. Haven't you been taught about numerical overflow?
Dana Ucaed wrote:if I have
Outputs are:
2541865828329
754810903
It may seem like this output is "wrong" or an error, but like Campbell said, it's numeric overflow. The compiler won't complain about it.
Numeric overflow happens because of the way integers are represented in bits. If you have a signed integer (one that can become negative) you need a "place" to hold these numbers. With bits you just have 0 and 1 to work with, no minus sign. So the decision was made to use "two's complement" to hold negatives. Two's complement says to invert the bits and add one to the result. It's probably easier to show an example. Assume we have eight bits (a byte). Then the number one looks like this: Two's complement first says to invert the bits: then add one With a byte, you can represent the numbers 127 to 128 this way. But notice how the representation for 1 looks like the number 256. If the byte is unsigned, this is the case. But with a signed byte, the numeric progression goes like this:
This is why you get negative numbers if a signed integer overflows.
All things are lawful, but not all things are profitable.
I presume that is the transverse product of consecutive digits, rather than the String whose digits you want to multiply. Otherwise the transverse product would be 0.Dana Ucaed wrote:. . . Output is:2091059712 . . . . why don't appear overflow when I used int as type of variable. . . .
Compare that number with the limits on the int type, and I think you should be able to see the explanation. You happened to be lucky with that part of the text.
And I have just seen a shortcut which you can use to reduce the amount of calculation you will have to do on Project Euler.
Campbell Ritchie wrote:2091059712's prime factors are 2⁹ × 3⁵ × 7⁵, which suggests that you have 7s, 3/6/9s and 2/4/8s in the text you are multiplying, and no 5/0s. Five 7s, but I can't tell how many of the other digits. It is very unlikely that you got that number from an overflow. Please show us the input which produced that output.
I have 2 posts because there are different implementations.
Original problem was the same but solutions are different.
OCAJP 7
Campbell Ritchie wrote:2091059712's prime factors are 2⁹ × 3⁵ × 7⁵, which suggests that you have 7s, 3/6/9s and 2/4/8s in the text you are multiplying, and no 5/0s. Five 7s, but I can't tell how many of the other digits. It is very unlikely that you got that number from an overflow. Please show us the input which produced that output.
9781797784617
(in the code I gave, use this:
@Dana,
with Project Euler, when brute force is a viable solution (by far not always applicable), it is always nice to look out for some optimizations. For instance, say:
S0 = d1 * d2 * d3 * ... * d13
S1 = d2 * d3 * d4 * ... * d13 * d14
can you write: S1 = S0 * alpha? And thus: what is alpha? And what if alpha < 1? Or > 1? And what has this to do with the excersize?
 1
Two 9s and a 6: there's the 3⁵ if we divide by 2.Piet Souris wrote:. . . 9781797784617 . . .
Five 7s: There's the 7⁵
Two 8s, a 4 and a 6: if we divide that by 3 we get 2⁹
Spot on It just so happens that this number will fit into the limits of an int without overflow errors.
Notice you have two 1s, which count as a unit for multiplication. Remembering what the products of other digits would be, work out what would be an annihilator. When you find that annihilator, you can move on to the next thirteen characters following it. I think PS is hinting about the same thing.
Dana Ucaed wrote:I know limits of datatype int.
Can someone say me the reason for which Output is:2091059712??
That is the largest product that your program could find that DIDN'T OVERFLOW. Other products (including the correct answer) did overflow and ended up being negative numbers which were less than maximumProduct.
The most difficult part of wild life photography is getting them to sign the model release.
Campbell Ritchie wrote:
Two 9s and a 6: there's the 3⁵ if we divide by 2.Piet Souris wrote:. . . 9781797784617 . . .
Five 7s: There's the 7⁵
Two 8s, a 4 and a 6: if we divide that by 3 we get 2⁹
Spot on It just so happens that this number will fit into the limits of an int without overflow errors.
Notice you have two 1s, which count as a unit for multiplication. Remembering what the products of other digits would be, work out what would be an annihilator. When you find that annihilator, you can move on to the next thirteen characters following it. I think PS is hinting about the same thing.
Carey Brown said it is overflow, but Campbell said it's not overflow.
What is the truth?
OCAJP 7
Piet Souris wrote:
Campbell Ritchie wrote:2091059712's prime factors are 2⁹ × 3⁵ × 7⁵, which suggests that you have 7s, 3/6/9s and 2/4/8s in the text you are multiplying, and no 5/0s. Five 7s, but I can't tell how many of the other digits. It is very unlikely that you got that number from an overflow. Please show us the input which produced that output.
9781797784617
(in the code I gave, use this:
This code is for Java 8?
I used Java 7.
@Dana,
with Project Euler, when brute force is a viable solution (by far not always applicable), it is always nice to look out for some optimizations. For instance, say:
S0 = d1 * d2 * d3 * ... * d13
S1 = d2 * d3 * d4 * ... * d13 * d14
can you write: S1 = S0 * alpha? And thus: what is alpha? And what if alpha < 1? Or > 1? And what has this to do with the excersize?
This exercise is for learning.
OCAJP 7
Dana Ucaed wrote:
Carey Brown said it is overflow, but Campbell said it's not overflow.
What is the truth?
They're both right, but Carey said it more clearly:
That is the largest product that your program could find that DIDN'T OVERFLOW. Other products (including the correct answer) did overflow and ended up being negative numbers which were less than maximumProduct.
All things are lawful, but not all things are profitable.
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