/* Triangle, square, pentagonal, hexagonal, heptagonal, and octagonal numbers are all figurate (polygonal) numbers and are generated by the following formulae: Triangle P3,n=n(n+1)/2 1, 3, 6, 10, 15, ... Square P4,n=n2 1, 4, 9, 16, 25, ... Pentagonal P5,n=n(3n−1)/2 1, 5, 12, 22, 35, ... Hexagonal P6,n=n(2n−1) 1, 6, 15, 28, 45, ... Heptagonal P7,n=n(5n−3)/2 1, 7, 18, 34, 55, ... Octagonal P8,n=n(3n−2) 1, 8, 21, 40, 65, ... The ordered set of three 4-digit numbers: 8128, 2882, 8281, has three interesting properties. The set is cyclic, in that the last two digits of each number is the first two digits of the next number (including the last number with the first). Each polygonal type: triangle (P3,127=8128), square (P4,91=8281), and pentagonal (P5,44=2882), is represented by a different number in the set. This is the only set of 4-digit numbers with this property. Find the sum of the only ordered set of six cyclic 4-digit numbers for which each polygonal type: triangle, square, pentagonal, hexagonal, heptagonal, and octagonal, is represented by a different number in the set. */ #include #include #include #include #include #include using namespace std; uint triangle(uint n){ return n * (n + 1) / 2; } uint square(uint n){ return n * n; } uint pentagonal(uint n){ return n * (3 * n - 1) / 2; } uint hexagonal(uint n){ return n * (2 * n - 1); } uint heptagonal(uint n){ return n * (5 * n - 3) / 2; } uint octogonal(uint n){ return n * (3 * n - 2); } bool validEnd(uint n){ return n % 100 >= 10; } std::set results; const unsigned int Limit = 10000; std::vector all(Limit, 0); uint finalMask = 0b111111000; void fillVector(vector & toFill, const uint bitmask, uint (*calc)(uint n)){ uint value = calc(0); for(uint i = 0; value < 10000; ++i){ if(value >= 1000 && validEnd(value)){ toFill[value] |= bitmask & finalMask; } value = calc(i + 1); } } void search(vector & sequence, uint mask = 0){ unsigned int from = 1000; unsigned int to = 10000; if(!sequence.empty()){ auto lowerTwoDigits = sequence.back() % 100; from = lowerTwoDigits * 100; to = from + 100; } for(auto next = from; next < to; ++next){ auto categories = all[next]; if(categories == 0){ continue; } bool isUnique = true; for(auto x : sequence){ if(x == next){ isUnique = false; break; } if(!isUnique){ continue; } for(auto j = 3; j <=8; ++j){ auto thisCategory = 1 << j; if((categories && thisCategory) == 0){ continue; } if((mask & thisCategory) != 0){ continue; } auto nextMask = mask | thisCategory; if(nextMask == finalMask){ auto first = sequence.front(); auto lowerTwoDigits = next % 100; auto upperTwoDigits = first / 100; if(lowerTwoDigits == upperTwoDigits){ auto sum = next; for(auto x : sequence){ sum += x; } results.insert(sum); } } else{ sequence.push_back(next); search(sequence, nextMask); sequence.pop_back(); } } } } } int main(){ cout << "Hello this is Patrick" << endl; auto start = chrono::high_resolution_clock::now(); // Finally learned what exactly it means when 1 << n, which is nice fillVector(all, 1 << 3, triangle); fillVector(all, 1 << 4, square); fillVector(all, 1 << 5, pentagonal); fillVector(all, 1 << 6, hexagonal); fillVector(all, 1 << 7, heptagonal); fillVector(all, 1 << 8, octogonal); vector sequence; search(sequence); for(auto x : results){ cout << x << endl; } auto duration = chrono::duration_cast(chrono::high_resolution_clock::now() - start); cout << (float)duration.count()/1000 << endl; return 0; }