S-tree (UVa 712) Two-pronged tree

Title: Https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=839&page=show_ problem&problem=653

Idea: The current node is K, left is 2k, right walk 2k + 1, get the last point, minus all the nodes in the front n layer, the corresponding 0 or 1 of the subscript is the answer.

/*S-tree (UVa 712)*/#include<iostream>#include<cstring>#include<vector>using namespacestd;Const intMAXN =Ten;intN;Chars[1&LT;&LT;MAXN], INPUT[MAXN];//Input and QueryintIDX[MAXN];//each layer corresponds to the variable XIvector<Char> output;//Output Query Resultsintsolve ();intMain () {//freopen ("Input.txt", "R", stdin);    intCNT =0;  while(CIN >> n && N! =0) {memset (s+1,0,sizeof(s+1)); memset (IDX,0,sizeof(IDX));                Output.clear ();  for(intI=1; i<=n; i++){            Charstr[3]; CIN>>str; Idx[i]= str[1] -'0'; }                intm; CIN>> s+1>>m;  for(intI=1; i<=m; i++) {cin>> input+1;        Output.push_back (S[solve ()); } cout<<"S-tree #"<< ++cnt <<":"<<Endl;  for(intI=0; i<m; i++) cout<<Output[i]; cout<< Endl <<Endl; }}intsolve () {intK =1;  for(intI=1; I<=strlen (input+1); i++){        intt = Input[idx[i]]-'0'; if(T = =0) K=2*K; Elsek=2*k +1; }    //cout << K-(1<<n) + 1 << endl;    return(K-(1<<n) +1); }

S-tree (UVa 712) Two-pronged tree