UVA-11925 Generating permutations Reasoning

The title of the topic is wrong, his second operation is to transfer the last one to the first, and then output to reverse the output

Problem-solving ideas: Similar bubble sorting method. First, if the first is N, then the back to the front, and then compare, take the largest put in front, this next time N to the front, n-1 on the back of N.

#include <cstdio>#include <vector>#include <deque>#define MAXN 310using namespace STD;intNUM[MAXN], N; deque<int>dq vector<int>AnsBOOLJudge () { for(inti =1; I <= N; i++)if(Dq[i-1]! = i)return false;return true;}intMain () { while(scanf("%d", &n) = =1&& N) {dq.clear (); Ans.clear ();intT for(inti =1; I <= N; i++) {scanf("%d", &t);        Dq.push_back (t); }intMAX =2* n * N; for(inti =0; I <= MAX; i++) {intx = dq[0];inty = dq[1];if(x = =1&& judge ()) { Break; }if(x > y && x! = N) {Ans.push_back (1); Swap (dq[0],dq[1]); }Else{Ans.push_back (2);                t = Dq.back ();                Dq.pop_back ();            Dq.push_front (t); }        } for(inti = ans.size ()-1; I >=0; i--)printf("%d", Ans[i]);printf("\ n"); }return 0;}

UVA-11925 Generating permutations Reasoning