You are given an array A of K integers where Ai denotes page number of a book. To compute the score, you can either add or multiply the last digit of the page numbers.
You have to find the maximum score you can get. Since the score can be quite large, output the score modulo 1000000007
Note: The book contains N pages. Also, you need to follow the order in which the page numbers are given in the array. Initially, your score is 0.
Input format :
First line: Two space seperated integers N and K.
Next line: K space seperated integers denoting the page numbers.
Output format :
Output the maximum score you can get. Print it modulo 1000000007
Input Constraints:
1<=N<=10^9
1<=k<=10^9
SAMPLE INPUT:
50 3
2 35 23
SAMPLE OUTPUT:
30
Explanation
Last digit of all page numbers are: 2, 5, and 3.
Initial score = 0
We add 2 to the score which now becomes 2, multiply with 5 making the score 10, finally we multiply with 3 making the score 30 which is the maximum score.
Output 30 % (10^9+7) = 30.
I encountered the same question in an online test I gave recently.
Instead N was the no of books and K is the size of the array.Both were given as inputs.
Here is what I did:
int main() {
long long n, k;
long long m = 1000000007;
cin >> n >> k;
vector<int> arr(k, 0);
for(int i = 0; i < k; i++){
cin >> arr[i];
}
long long prod = 1;
long long sum = 0;
for(int i = 0; i < k; i++){
if(arr[k] < n){
prod = ((prod % m) * (arr[k] % 10)) % m;
sum = ((sum% m) + (arr[k] % 10)) % m;
prod = max(prod, sum);
sum = max(prod, sum);
}
}
cout << prod % m << endl;
return 0;
}
As you can see instead of handling for 1 and 2, I am checking for max value of the product and sum at every iteration and updating both the product and sum with it.
I got two test cases passed and rest gave wrong answer.Why is it so?
Here is the question link, if anyone needs to give it a try.
The Book Game Problem
The problem asks you to add OR multiply the last digit of the page numbers to make the resultant score as large as possible.
In this case, you should add when the digit is 0 or 1, and multiply otherwise.
For example,
Let the last digit sequence be
[1 0 2 5 8 1]
'score' is initialized to be 0.
add 1 (score: 1)
add 0 (score: 1)
multiply by 2 (score: 2)
multiply by 5 (score: 10)
multiply by 8 (score: 80)
add 1 (score: 81)
and before submitting the answer, you need to modulo it by 1000000007.
so,
score %= 1000000007
in your code, you are calculating the prod and the sum separately, which is not what you are asked to do. Also, you are sumbitting only the 'prod' value, while it is not always the maximum value (consider multiplying some number by 0)
And additionally, you are modulo-ing the intermediate values (prod and sum) which can lead to wrong answer. The modulo should not be used to calculate the score, but to truncate the result of the score digits.
So, my answer is,
Calculate the intermediate values and assign it to one variable named 'score' (don't separate the product and the sum), and modulo the value just before printing the output (don't modulo it every time you add or multiply)
Thanks.
Related
I have been given the following assignment:
Given N integers in the form of A(i) where 1≤i≤N, make each number
A(i) in the N numbers equal to M. To convert a number A(i) to M, it
will cost |M−Ai| units. Find out the minimum cost to convert all the N
numbers to M, so you should choose the best M to get the minimum cost.
Given:
1 <= N <= 10^5
1 <= A(i) <= 10^9
My approach was to calculate the sum of all numbers and find avg = sum / n and then subtract each number by avg to get the minimum cost.
But this fails in many test cases. How can I find the optimal solution for this?
You should take the median of the numbers (or either of the two numbers nearest the middle if the list has even length), not the mean.
An example where the mean fails to minimize is: [1, 2, 3, 4, 100]. The mean is 110 / 5 = 22, and the total cost is 21 + 20 + 19 + 18 + 78 = 156. Choosing the median (3) gives total cost: 2 + 1 + 0 + 1 + 97 = 101.
An example where the median lies between two items in the list is [1, 2, 3, 4, 5, 100]. Here the median is 3.5, and it's ok to either use M=3 or M=4. For M=3, the total cost is 2 + 1 + 0 + 1 + 2 + 97 = 103. For M=4, the total cost is 3 + 2 + 1 + 0 + 1 + 96 = 103.
A formal proof of correctness can be found on Mathematics SE, although you may convince yourself of the result by noting that if you nudge M a small amount delta in one direction (but not past one of the data points) -- and for example's sake let's say it's in the positive direction, the total cost increases by delta times the number of points to the left of M minus delta times the number of points to the right of M. So M is minimized when the number of points to its left and the right are equal in number, otherwise you could move it a small amount one way or the other to decrease the total cost.
#PaulHankin already provided a perfect answer. Anyway, when thinking about the problem, I didn't think of the median being the solution. But even if you don't know about the median, you can come up with a programming solution.
I made similar observations as #PaulHankin in the last paragraph of his last answer. This made me realize, that I have to eliminate outliers iteratively in order to find m. So I wrote a program that first sorts the input array (vector) A and then analyzes the minimum and maximum values.
The idea is to move the minimum values towards the second smallest values and the maximum values towards the second largest values. You always move either the minimum or maximum values, depending on whether you have less minimum values than maximum values or not. If all array items end up being the same value, then you found your m:
#include <vector>
#include <algorithm>
#include <iostream>
using namespace std;
int getMinCount(vector<int>& A);
int getMaxCount(vector<int>& A);
int main()
{
// Example as given by #PaulHankin
vector<int> A;
A.push_back(1);
A.push_back(2);
A.push_back(3);
A.push_back(4);
A.push_back(100);
sort(A.begin(), A.end());
int minCount = getMinCount(A);
int maxCount = getMaxCount(A);
while (minCount != A.size() && maxCount != A.size())
{
if(minCount <= maxCount)
{
for(int i = 0; i < minCount; i++)
A[i] = A[minCount];
// Recalculate the count of the minium value, because we changed the minimum.
minCount = getMinCount(A);
}
else
{
for(int i = 0; i < maxCount; i++)
A[A.size() - 1 - i] = A[A.size() - 1 - maxCount];
// Recalculate the count of the maximum value, because we changed the maximum.
maxCount = getMaxCount(A);
}
}
// Print out the one and only remaining value, which is m.
cout << A[0] << endl;
return 0;
}
int getMinCount(vector<int>& A)
{
// Count how often the minimum value exists.
int minCount = 1;
int pos = 1;
while (pos < A.size() && A[pos++] == A[0])
minCount++;
return minCount;
}
int getMaxCount(vector<int>& A)
{
// Count how often the maximum value exists.
int maxCount = 1;
int pos = A.size() - 2;
while (pos >= 0 && A[pos--] == A[A.size() - 1])
maxCount++;
return maxCount;
}
If you think about the algorithm, then you will come to the conclusion, that it actually calculates the median of the values in the array A. As example input I took the first example given by #PaulHankin. As expected, the code provides the correct result (3) for it.
I hope my approach helps you to understand how to tackle such kind of problems even if you don't know the correct solution. This is especially helpful when you are in an interview, for example.
This is my problem.
Given an array of integers and another integer k, find the sum of differences of each element of the array and k.
For example if the array is 2, 4, 6, 8, 10 and k is 3
Sum of difference
= abs(2 - 3) + abs(4-3) + abs(6 - 3) + abs(8 - 3) + abs(10 - 3)
= 1 + 1 + 3 + 5 + 7
= 17
The array remains the same throughout and can contain up to 100000 elements and there will be 100000 different values of k to be tested. k may or may not be an element of the array. This has to be done within 1s or about 100M operations. How do I achieve this?
You can run multiple queries for sums of absolute differences in O(log N) if you add a preprocessing step which costs O(N * log N).
Sort the array, then for each item in the array store the sum of all numbers that are smaller than or equal to the corresponding item. This can be done in O(N * log N) Now you have a pair of arrays that look like this:
2 4 6 8 10 // <<== Original data
2 6 12 20 30 // <<== Partial sums
In addition, store the total T of all numbers in the array.
Now you can get sums of absolute differences by running a binary search on the original array, and using the sums from the partial sums array to compute the answer: subtract the sum of all numbers to the left of the target k from the count of numbers to the left of the target times k, then subtract the count times k from the sum to the right of the number, and add the two numbers together. The partial sum of the numbers to the right of the number can be computed by subtracting the partial sum on the left from the total T.
For k=3 binary search gets you to position 1.
Partial sum on the left is 2
Count of items on the left is 1
Partial sum on the right is (30-2)=28
Count of items on the right is 4
You compute (1*3-2) + (28-4*3) = 1 + 16 = 17
First sort the array and then compute an array that stores the sum of the prefixes of the resulting sorted array. Let's denote this array p, you can compute p in linear time so that p[i] = a[0] + a[1] + ... a[i]. Now having this array you can answer with constant complexity the question what is the sum of elements a[x] + a[x+1] + .... +a[y](i.e. with indices x to y). To do that you simply compute p[y] - p[x-1](Take special care when x is 1).
Now to answer a query of the type what is the sum of absolute differences with k, we will split the problem in two parts - what is the sum of the numbers greater than k and the numbers smaller than k. In order to compute these, perform a binary search to find the position of k in the sorted a(denote that idx), and compute the sum of the values in a before idx(denote that s) and after idx(denote that S). Now the sum of absolute differences with k is idx * k - s + S - (a.length - idx)* k. This of course is pseudo code and what I mean by a.length is the number of elements in a.
After performing a linearithmic precomputation, you will be able to answer a query with O(log(n)). Please note this approach only makes sense if you plan to perform multiple queries. If you are only going to perform a single query, you can not possibly go faster than O(n).
Just implementing dasblinkenlight's solution in "contest C++":
It does exactly as he says. Reads the values, sorts them, stores the accumulated sum in V[i].second, but here V[i] is the acumulated sum until i-1 (to simplify the algorithm). It also stores a sentinel in V[n] for cases when the query is greater than max(V).
Then, for each query, binary search for the value. In this case V[a].second is the sum of values lesser than query, V[n].second-V[a].second is the sum of values greater than it.
#include<iostream>
#include<algorithm>
#define pii pair<int, int>
using namespace std;
pii V[100001];
int main() {
int n;
while(cin >> n) {
for(int i=0; i<n; i++)
cin >> V[i].first;
sort(V, V+n);
V[0].second = 0;
for(int i=1; i<=n; i++)
V[i].second = V[i-1].first + V[i-1].second;
int k; cin >> k;
for(int i=0; i<k; i++) {
int query; cin >> query;
pii* res = upper_bound(V, V+n, pii(query, 0));
int a = res-V, b=n-(res-V);
int left = query*a-V[a].second;
int right = V[n].second-V[a].second-query*b;
cout << left+right << endl;
}
}
}
It assumes a file with a format like this:
5
10 2 8 4 6
2
3 5
Then, for each query, it answers like this:
17
13
I am given the set {1, 2, 3, ... ,N}. I have to find the maximum size of a subset of the given set so that the sum of any 2 numbers from the subset is not divisible by a given number K. N and K can be up to 2*10^9 so i need a very fast algorithm. I only came up with an algorithm of complexity O(K), which is slow.
first calculate all of the set elements mod k.and solve simple problem:
find the maximum size of a subset of the given set so that the sum of any 2 numbers from the subset is not equal by a given number K.
i divide this set to two sets (i and k-i) that you can not choose set(i) and set(k-i) Simultaneously.
int myset[]
int modclass[k]
for(int i=0; i< size of myset ;i++)
{
modclass[(myset[i] mod k)] ++;
}
choose
for(int i=0; i< k/2 ;i++)
{
if (modclass[i] > modclass[k-i])
{
choose all of the set elements that the element mod k equal i
}
else
{
choose all of the set elements that the element mod k equal k-i
}
}
finally you can add one element from that the element mod k equal 0 or k/2.
this solution with an algorithm of complexity O(K).
you can improve this idea with dynamic array:
for(int i=0; i< size of myset ;i++)
{
x= myset[i] mod k;
set=false;
for(int j=0; j< size of newset ;j++)
{
if(newset[j][1]==x or newset[j][2]==x)
{
if (x < k/2)
{
newset[j][1]++;
set=true;
}
else
{
newset[j][2]++;
set=true;
}
}
}
if(set==false)
{
if (x < k/2)
{
newset.add(1,0);
}
else
{
newset.add(0,1);
}
}
}
now you can choose with an algorithm of complexity O(myset.count).and your algorithm is more than O(myset.count) because you need O(myset.count) for read your set.
complexity of this solution is O(myset.count^2),that you can choose algorithm depended your input.with compare between O(myset.count^2) and o(k).
and for better solution you can sort myset based on mod k.
I'm assuming that the set of numbers is always 1 through N for some N.
Consider the first N-(N mod K) numbers. The form floor(N/K) sequences of K consecutive numbers, with reductions mod K from 0 through K-1. For each group, floor(K/2) have to be dropped for having a reduction mod K that is the negation mod K of another subset of floor(K/2). You can keep ceiling(K/2) from each set of K consecutive numbers.
Now consider the remaining N mod K numbers. They have reductions mod K starting at 1. I have not worked out the exact limits, but if N mod K is less than about K/2 you will be able to keep all of them. If not, you will be able to keep about the first ceiling(K/2) of them.
==========================================================================
I believe the concept here is correct, but I have not yet worked out all the details.
==========================================================================
Here is my analysis of the problem and answer. In what follows |x| is floor(x). This solution is similar to the one in #Constantine's answer, but differs in a few cases.
Consider the first K*|N/K| elements. They consist of |N/K| repeats of the reductions modulo K.
In general, we can include |N/K| elements that are k modulo K subject to the following limits:
If (k+k)%K is zero, we can include only one element that is k modulo K. That is the case for k=0 and k=(K/2)%K, which can only happen for even K.
That means we get |N/K| * |(K-1)/2| elements from the repeats.
We need to correct for the omitted elements. If N >= K we need to add 1 for the 0 mod K elements. If K is even and N>=K/2 we also need to add 1 for the (K/2)%K elements.
Finally, if M(N)!=0 we need to add a partial or complete copy of the repeat elements, min(N%K,|(K-1)/2|).
The final formula is:
|N/K| * |(K-1)/2| +
(N>=K ? 1 : 0) +
((N>=K/2 && (K%2)==0) ? 1 : 0) +
min(N%K,|(K-1)/2|)
This differs from #Constantine's version in some cases involving even K. For example, consider N=4, K=6. The correct answer is 3, the size of the set {1, 2, 3}. #Constantine's formula gives |(6-1)/2| = |5/2| = 2. The formula above gets 0 for each of the first two lines, 1 from the third line, and 2 from the final line, giving the correct answer.
formula is
|N/K| * |(K-1)/2| + ost
ost =
if n<k:
ost =0
else if n%k ==0 :
ost =1
else if n%k < |(K-1)/2| :
ost = n%k
else:
ost = |(K-1)/2|
where |a/b|
for example |9/2| = 4 |7/2| = 3
example n = 30 , k =7 ;
1 2 3 4 5 6 7
8 9 10 11 12 13 14
15 16 17 18 19 20 21
22 23 24 25 26 27 28
29 30
1 2 3 |4| 5 6 7. - is first line .
8 9 10 |11| 12 13 14 - second line
if we getting first 3 number in each line we may get size of this subset. also we may adding one number from ( 7 14 28)
getting first 3 number (1 2 3) is a number |(k-1)/2| .
a number of this line is |n/k| .
if there is not residue we may add one number (for example last number).
if residue < |(k-1)/2| we get all number in last line
else getting |(K-1)/2|.
thanks for exception case.
ost = 0 if k>n
n,k=(raw_input().split(' '))
n=int(n)
k=int(k)
l=[0 for x in range(k)]
d=[int(x) for x in raw_input().split(' ')]
flag=0
for x in d:
l[x%k]=l[x%k]+1
sum=0
if l[0]!=0:
sum+=1
if (k%2==0):
sum+=1
if k==1:
print 1
elif k==2:
print 2
else:
i=1
j=k-1
while i<j:
sum=sum+(l[i] if l[i]>=l[j] else l[j])
i=i+1
j=j-1
print sum
This is explanation to ABRAR TYAGI and amin k's solution.
The approach to this solution is:
Create an array L with K buckets and group all the elements from the
input array D into the K buckets. Each bucket L[i] contains D's elements such that ( element % K ) = i.
All the elements that are individually divisible by K are in L[0]. So
only one of these elements (if any) can belong in our final (maximal)
subset. Sum of any two of these elements is divisible by K.
If we add an element from L[i] to an element in L[K-i] then the sum is divisible by K. Hence we can add elements from only one of these buckets to
our final set. We pick the largest bucket.
Code:
d is the array containing the initial set of numbers of size n. The goal of this code is to find the count of the largest subset of d such that the sum of no two integers is divisible by 2.
l is an array that will contain k integers. The idea is to reduce each (element) in array d to (element % k) and save the frequency of their occurrences in array l.
For example, l[1] contains the frequency of all elements % k = 1
We know that 1 + (k-1) % k = 0 so either l[1] or l[k-1] have to be discarded to meet the criteria that sum of no two numbers % k should be 0.
But as we need the largest subset of d, we choose the larger of l[1] and l[k-1]
We loop through array l such that for (i=1; i<=k/2 && i < k-i; i++) and do the above step.
There are two outliers. The sum of any two numbers in the l[0] group % k = 0. So add 1 if l[0] is non-zero.
if k is even, the loop does not handle i=k/2, and using the same logic as above increment the count by one.
Problem: "An algorithm to find the number of six digit numbers where the sum of the first three digits is equal to the sum of the last three digits."
I came across this problem in an interview and want to know the best solution. This is what I have till now.
Approach 1: The Brute force solution is, of course, to check for each number (between 100,000 and 999,999) whether the sum of its first three and last three digits are equal. If yes, then increment certain counter which keeps count of all such numbers.
But this checks for all 900,000 numbers and so is inefficient.
Approach 2: Since we are asked "how many" such numbers and not "which numbers", we could do better. Divide the number into two parts: First three digits (these go from 100 to 999) and Last three digits (these go from 000 to 999). Thus, the sum of three digits in either part of a candidate number can range from 1 to 27.
* Maintain a std::map<int, int> for each part where key is the sum and value is number of numbers (3 digit) having that sum in the corresponding part.
* Now, for each number in the first part find out its sum and update the corresponding map.
* Similarly, we can get updated map for the second part.
* Now by multiplying the corresponding pairs (e.g. value in map 1 of key 4 and value in map 2 of key 4) and adding them up we get the answer.
In this approach, we end up checking 1K numbers.
My question is how could we further optimize? Is there a better solution?
For 0 <= s <= 18, there are exactly 10 - |s - 9| ways to obtain s as the sum of two digits.
So, for the first part
int first[28] = {0};
for(int s = 0; s <= 18; ++s) {
int c = 10 - (s < 9 ? (9 - s) : (s - 9));
for(int d = 1; d <= 9; ++d) {
first[s+d] += c;
}
}
That's 19*9 = 171 iterations, for the second half, do it similarly, with the inner loop starting at 0 instead of 1, that's 19*10 = 190 iterations. Then sum first[i]*second[i] for 1 <= i <= 27.
Generate all three-digit numbers; partition them into sets based on their sum of digits. (Actually, all you need to do is keep a vector that counts the size of the sets). For each set, the number of six-digit numbers that can be generated is the size of the set squared. Sum up the squares of the set sizes to get your answer.
int sumCounts[28]; // sums can go from 0 through 27
for (int i = 0; i < 1000; ++i) {
sumCounts[sumOfDigits(i)]++;
}
int total = 0;
for (int i = 0; i < 28; ++i) {
count = sumCounts[i];
total += count * count;
}
EDIT Variation to eliminate counting leading zeroes:
int sumCounts[28];
int sumCounts2[28];
for (int i = 0; i < 100; ++i) {
int s = sumOfDigits(i);
sumCounts[s]++;
sumCounts2[s]++;
}
for (int i = 100; i < 1000; ++i) {
sumCounts[sumOfDigits(i)]++;
}
int total = 0;
for (int i = 0; i < 28; ++i) {
count = sumCounts[i];
total += (count - sumCounts2[i]) * count;
}
Python Implementation
def equal_digit_sums():
dists = {}
for i in range(1000):
digits = [int(d) for d in str(i)]
dsum = sum(digits)
if dsum not in dists:
dists[dsum] = [0,0]
dists[dsum][0 if len(digits) == 3 else 1] += 1
def prod(dsum):
t = dists[dsum]
return (t[0]+t[1])*t[0]
return sum(prod(dsum) for dsum in dists)
print(equal_digit_sums())
Result: 50412
One idea: For each number from 0 to 27, count the number of three-digit numbers that have that digit sum. This should be doable efficiently with a DP-style approach.
Now you just sum the squares of the results, since for each answer, you can make a six-digit number with one of those on each side.
Assuming leading 0's aren't allowed, you want to calculate how many different ways are there to sum to n with 3 digits. To calculate that you can have a for loop inside a for loop. So:
firstHalf = 0
for i in xrange(max(1,n/3),min(9,n+1)): #first digit
for j in xrange((n-i)/2,min(9,n-i+1)): #second digit
firstHalf +=1 #Will only be one possible third digit
secondHalf = firstHalf + max(0,10-|n-9|)
If you are trying to sum to a number, then the last number is always uniquely determined. Thus in the case where the first number is 0 we are just calculating how many different values are possible for the second number. This will be n+1 if n is less than 10. If n is greater, up until 18 it will be 19-n. Over 18 there are no ways to form the sum.
If you loop over all n, 1 through 27, you will have your total sum.
This question already has answers here:
Closed 11 years ago.
Possible Duplicate:
Generating the partitions of a number
Prime number sum
The number 7 can be expressed in 5 ways as a sum of primes:
2 + 2 + 3
2 + 3 + 2
2 + 5
3 + 2 + 2
5 + 2
Make a program that calculates, in how many ways number n can be
expressed as a sum of primes. You can assume that n is a number
between 0-100. Your program should print the answer in less than a
second
Example 1:
Give number: 7 Result: 5
Example 2:
Give number: 20 Result: 732
Example 3:
Give number: 80 Result: 10343662267187
I've been at this problem for hours. I can't figure out how to get n from (n-1).
Here are the sums from the first 30 numbers by a tree search
0 0 0 1 2 2 5 6 10 16 19 35 45 72 105 152 231 332 500 732 1081 1604 2351 3493 5136 7595 11212 16534 24441
I thought I had something with finding the biggest chain 7 = 5+2 and somehow using the knowledge that five can be written as 5, 3+2, 2+3, but somehow I need to account for the duplicate 2+3+2 replacement.
Look up dynamic programming, specifically Wikipedia's page and the examples there for the fibonacci sequence, and think about how you might be able to adapt that to your problem here.
Okay so this is a complicated problem. you are asking how to write code for the Partition Function; I suggest that you read up on the partition function itself first. Next you should look at algorithms to calculate partitions. It is a complex subject here is a starting point ... Partition problem is [NP complete] --- This question has already been asked and answered here and that may also help you start with algorithms.
There're several options. Since you know the number is between 0-100, there is the obvious: cheat, simply make an array and fill in the numbers.
The other way would be a loop. You'd need all the primes under 100, because a number which is smaller than 100 can't be expressed using the sum of a prime which is larger than 100. Eg. 99 can't be expressed as the sum of 2 and any prime larger than 100.
What you also know is: the maximum length of the sum for even numbers is the number divided by 2. Since 2 is the smallest prime. For odd numbers the maximum length is (number - 1) / 2.
Eg.
8 = 2 + 2 + 2 + 2, thus length of the sum is 4
9 = 2 + 2 + 2 + 3, thus length of the sum is 4
If you want performance you could cheat in another way by using GPGPU, which would significantly increase performance.
Then they're is the shuffling method. If you know 7 = 2 + 2 + 3, you know 7 = 2 + 3 + 2. To do this you'd need a method of calculating the different possibilities of shuffling. You could store the combinations of possibilities or keep them in mind while writing your loop.
Here is a relative brute force method (in Java):
int[] primes = new int[]{/* fill with primes < 100 */};
int number = 7; //Normally determined by user
int maxLength = (number % 2 == 0) ? number / 2 : (number - 1) / 2; //If even number maxLength = number / 2, if odd, maxLength = (number - 1) / 2
int possibilities = 0;
for (int i = 1; i <= maxLength; i++){
int[][] numbers = new int[i][Math.pow(primes.length, i)]; //Create an array which will hold all combinations for this length
for (int j = 0; j < Math.pow(primes.length, i); j++){ //Loop through all the possibilities
int value = 0; //Value for calculating the numbers making up the sum
for (int k = 0; k < i; k++){
numbers[k][j] = primes[(j - value) % (Math.pow(primes.length, k))]; //Setting the numbers making up the sum
value += numbers[k][j]; //Increasing the value
}
}
for (int x = 0; x < primes.length; x++){
int sum = 0;
for (int y = 0; y < i; y++){
sum += numbers[y];
if (sum > number) break; //The sum is greater than what we're trying to reach, break we've gone too far
}
if (sum == number) possibilities++;
}
}
I understand this is complicated. I will try to use an analogy. Think of it as a combination lock. You know the maximum number of wheels, which you have to try, hence the "i" loop. Next you go through each possibility ("j" loop) then you set the individual numbers ("k" loop). The code in the "k" loop is used to go from the current possibility (value of j) to the actual numbers. After you entered all combinations for this amount of wheels, you calculate if any were correct and if so, you increase the number of possibilities.
I apologize in advance if I made any errors in the code.