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Project Euler #32: Pandigital products (C++)

Question
We shall say that an n-digit number is pandigital if it makes use of all the digits 1 to n exactly once; for example, the 5-digit number, 15234, is 1 through 5 pandigital.
The product 7254 is unusual, as the identity, 39 × 186 = 7254, containing multiplicand, multiplier, and product is 1 through 9 pandigital.
Find the sum of all products whose multiplicand/multiplier/product identity can be written as a 1 through 9 pandigital.
HINT: Some products can be obtained in more than one way so be sure to only include it once in your sum.

Answer : 45228

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Solution


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#include <iostream>
#include <vector>
#include <set>
#include <algorithm>

int main() {
    unsigned int N;
    std::cin >> N;
    
    std::vector<unsigned int> digs = {1, 2, 3, 4, 5, 6, 7, 8, 9};
    digs.resize(N);
    std::set<unsigned int> val;
    do {
        for(unsigned int Aln = 1; Aln < N; Aln++) {
            for(unsigned int Bln = 1; Bln < N - Aln; Bln++) {
                unsigned int Cln = N - Aln - Bln;
                if(Cln < Aln || Cln < Bln) {
                    break;
                }
                unsigned int pos = 0;
                unsigned int a = 0;
                for(unsigned int i = 1; i <= Aln; i++) {
                    a *= 10;
                    a += digs[pos++];
                }
                unsigned int b = 0;
                for(unsigned int i = 1; i <= Bln; i++) {
                    b *= 10;
                    b += digs[pos++];
                }
                unsigned int c = 0;
                for(unsigned int i = 1; i <= Cln; i++) {
                    c *= 10;
                    c += digs[pos++];
                }
                if((a * b) == c) {
                    val.insert(c);
                }
            }
        }
    } while(std::next_permutation(digs.begin(), digs.end()));
    unsigned int sum = 0;
    for(auto x : val) {
        sum += x;
    }
    std::cout << sum << std::endl;
    return 0;
}
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