All possible sum from a given of coins - c++

Question

You have n coins with certain values. Your task is to find all the money sums you can create using these coins.
Input
The first input line has an integer n: the number of coins.
The next line has n integers x1,x2,…,xn: the values of the coins.
Output
First print an integer k: the number of distinct money sums. After this, print all possible sums in increasing order.
Constraints
1≤n≤100
1≤xi≤1000
Example
Input:
4
4 2 5 2
Output:
9
2 4 5 6 7 8 9 11 13
I have written a code which works perfectly for the small inputs but gives the wrong answer to the large inputs. Please help to find the mistake and how do I correct it.
my code is:
#include <bits/stdc++.h>
using namespace std;
set<long long> s;
// Prints sums of all subsets of array
void subsetSums(long long arr[], long long n)
{
// There are totoal 2^n subsets
long long total = 1 << n;
// Consider all numbers from 0 to 2^n - 1
for (long long i = 0; i < total; i++)
{
long long sum = 0;
// Consider binary reprsentation of
// current i to decide which elements
// to pick.
for (long long j = 0; j < n; j++)
if (i & (1 << j))
sum += arr[j];
// Print sum of picked elements.
if (sum)
s.insert(sum);
}
}
// Driver code
int main()
{
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
long long n;
cin >> n;
long long arr[n];
for (long long i = 0; i < n; i++)
{
cin >> arr[i];
}
subsetSums(arr, n);
cout << s.size() << "\n";
for (auto it = s.begin(); it != s.end(); ++it)
cout << *it << " ";
return 0;
}
for example, it gives the wrong answer for
50
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
as
18
2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36
the correct output should be:
50
1 2 3 4 ... 50

Answer 1

your code is simply too slow 2^n subsets gives ‭1,267,650,600,228,229,401,496,703,205,376‬ subsets in the worst case (when n=100) while C++ does on average about 1000,000,000 operations per second.
This problem can be solved with dynamic programming, consider having an array dp of size 100001, so that dp[x] denotes if sum of x is possible to achieve.
Base case is easy - sum of 0 is possible without using any coins: dp[0]=1
Then for each coin we can try to increase existing sums by coins value to fill up our table:
for each coinValue:
for coinSum = 100000 - coinValue; coinSum >=0; coinSum--)
if(dp[coinSum])
dp[coinSum + coinValue]=1
Notice that we are looping backwards, this is done on purpose so that each coin gets used only once.
Complexity: O(n^2*maxCoinValue)

Answer 2

Your algorithm is poor, but the reason you're getting wrong results is because you're overflowing int. long long total = 1<<n; shifts an int left by n places, and the fact you're assigning the result to a long long is irrelevant.
You can find problems like this using ubsan. Here's a reproduction of your problem, including warning messages from ubsan:
$ clang++ -fsanitize=undefined a.cpp -o a && ./a
50
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
a.cpp:11:25: runtime error: shift exponent 50 is too large for 32-bit type 'int'
a.cpp:22:24: runtime error: shift exponent 32 is too large for 32-bit type 'int'
18
2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36