Combined mathematical exercises (generated and arranged by Reverse columns)

/*************************************** * *********** Use the line segment tree to count the number of spaces, to determine where a number should be stored in the arrangement, the total time complexity is N * logn ********************************** * ******************/# include <iostream> # include <fstream> using namespace STD; # define maxn 1001 # define L (x) (x <1) # define R (x) (x <1) | 1) fstream fout ("out.txt ", IOS: Out); struct treenode {int L, R, space, P ;}; treenode node [maxn * 5]; void build_tree (int u, int L, int R) {node [u]. L = L; node [u]. R = r; node [u]. space = r-L + 1; node [u]. P = r; If (L = r) return; int mid = (L + r)> 1; build_tree (L (u), L, mid ); build_tree (R (u), Mid + 1, R);} int position (int u, int num) // determine the subscript, so that it is an empty position of num {node [u]. space --; If (node [u]. L = node [u]. r) return node [u]. l; If (Num <= node [L (u)]. space) return position (L (u), num); else return position (R (u), num-node [L (u)]. space);} void reordertopermu (int n, int * reorder, int * permu) // converts Reverse Order columns into arrays {for (INT I = 1; I <= N; I ++) {int Pos = position (1, reorder [I] + 1); permu [POS] = I ;}} int main () {int N; int reorder [maxn], permu [maxn]; while (CIN> N) // input the size of the reverse order column {build_tree (1, 1, n ); for (INT I = 1; I <= N; I ++) CIN> reorder [I]; reordertopermu (n, reorder, permu); For (INT I = 1; I <= N; I ++) fout <permu [I]; fout <Endl;} return 0 ;}