http://acm.cs.ecnu.edu.cn/problem.php?problemid=1443
题目的大意就是说密码是n位的，所以有10^n种情况，然后由于当输入密码长度大于n时，只有最后n为有效。这样我们只要用一个（10^n+n-1）长的数字序列，就能破解这个密码。
用DFS找一个欧拉回路，按照字典序输出路径。由于状态过多，不能用递归，要用非递归实现DFS，然后用stack会超时，那就自己写个数组吧
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cmath> 5 #include <algorithm> 6 #include <queue> 7 #include <stack> 8 #include <vector> 9 #include <string> 10 #include <set> 11 #include <cctype> 12 #include <cstdlib> 13 #include <map> 14 #define maxn 1000100 15 #define inf 0x3f3f3f 16 #define mod 9997 17 #define pi 3.1415926535898 18 #define LL long long 19 using namespace std; 20 21 int t[maxn], head[maxn], len[maxn], next[maxn], top, cnt, n, dk, ans[maxn]; 22 bool vis[maxn]; 23 int fc[8] = {1, 10, 100, 1000, 10000, 100000, 1000000}; 24 struct node{ 25 int x, p; 26 }st[maxn]; 27 28 void add(int u, int v, int w){ 29 t[cnt] = v; 30 len[cnt] = w; 31 next[cnt] = head[u]; 32 head[u] = cnt++; 33 } 34 35 void solve(){ 36 memset(head, -1, sizeof(head)); 37 memset(vis, 0, sizeof(vis)); 38 cnt = 0; 39 int tmp; 40 for (int i = 0; i < fc[n - 1]; i++){ 41 tmp = i % fc[n - 2]; 42 for (int j = 9; j >= 0; j--) 43 add(i, tmp * 10 + j, i * 10 + j); 44 } 45 int u, pos; 46 top = 2; dk = 0; 47 st[1].x = 0; st[1].p = head[0]; 48 while (top){ 49 u = st[top - 1].x; pos = st[top - 1].p; 50 for (; ~pos; pos = next[pos]) 51 if (!vis[pos]){ 52 st[top - 1].p = pos; 53 vis[pos] = true; 54 st[top].p = head[t[pos]]; st[top].x = t[pos]; top++; 55 break; 56 } 57 if (pos == -1){ 58 ans[++dk] = st[top - 1].p; 59 top--; 60 } 61 } 62 for (int i = 1; i < n; i++) printf("0"); 63 for (int i = dk - 1; i >= 2; i--) printf("%d", len[ans[i]] % 10); 64 printf("\n"); 65 } 66 67 int main(){ 68 while (cin >> n && n){ 69 if (n == 1){ 70 printf("0123456789\n"); 71 continue; 72 }else{ 73 solve(); 74 } 75 } 76 return 0; 77 }