/* by bell Rules for gem-fusion challenge by Anton Böhler • in total there are 8 diffrent gem kinds • in a fusion there are at most 6 gems involved • the gems entered are in an Array/List and not sorted • gems are represented as characters (r, o, y, g, b, p) • not every combination of gems yields a fusion (-> Table for fusions) Task: for a given List of gems in a fusion, output the obtained gems of that fusion. Table for fusions (you can create your own too if you want):roygbp r-> g+b 100000->000110 g+g -> r+r 000200->200000 r+r->o 200000->010000 b->g 000010->000100 r+o->r+y 110000->101000 4r->r+5o 400000->150000 2r+g+b+y+o ->b+p 211110->000011 p->3b+3r 000001->300030 2b+p->2p 000021->000002 g+y->b+y 001100->001010 2p->4y 000002->004000 6y->r+g+b+3p 006000->100113 🔶💎😇💍🔮⚗️🔻 */ #include #include #include using namespace std; int main() { srand(time(NULL)); int k,diff,difsum,d; vectorv(6,0); //frequencies for gems \U0001f48e vectorsum={1,2,2,1,2,4,6,1,3,2,2,6}; vectort={100000, 110, 200, 200000, 200000, 10000, 10, 100,110000,101000,400000,150000,211110,11,1,300030, 21,2,1100, 1010, 2,4000, 6000,100113}; vectorout={"r-> g+\U0001f48e","g+g -> r+r","r+r->o","\U0001f48e->g","r+o->r+y","4r->r+5o","2r+g+\U0001f48e+y+o ->\U0001f48e+p","p->3\U0001f48e+3r","2\U0001f48e+p->2p","g+y->2\U0001f48e+y","2p->4y","6y->r+g+\U0001f48e+3p "}; //vectoro={"r-> g+b","g+g -> r+r","r+r->o","b->g","r+o->r+y","4r->r+5o","2r+g+b+y+o ->b+p","p->3b+3r","2b+p->2p","g+y->b+y","2p->4y","6y->r+g+b+3p "}; const string g="roygbp"; string gems=""; //random gems for(int i=0;i<6;i++){ k=rand()%6; v[k]++; gems+=g[k]; } k=0; cout<<"The gems are "<=0&&(!(6-sum[n/2]-difsum))){ cout<=0&&(!(6-sum[n/2]-difsum))){ cout<

gemFusionAntonBohler