牛客多校(2020第十场)A Permutation

 

 

输入

2
3
5

输出

1 2
1 2 4 3

题解：

每次能*2就*2，不行就*3

 1 #include
 2 #include
 3 #include
 4 #include
 5 #include
 6 #include
 7 #include
 8 
 9 using namespace std;
10 #define ll long long 
11 const ll N = 1e6 + 5;
12 
13 inline ll read() {
14     ll x = 0, f = 1;
15     char ch = getchar();
16     while(ch<'0'||ch>'9'){
17         if(ch=='-')
18             f=-1;
19         ch=getchar();
20     }
21     while(ch>='0'&&ch<='9'){
22         x = x * 10 + ch - '0';
23         ch = getchar();
24     }
25     return x * f;
26 }
27 
28 bool visit[N];
29 vector <int> res;
30 
31 void solve() {
32     res.clear();
33     fill (visit, visit+N, false);
34     int p = read();
35     int cnt = 1, num = 1;
36     visit[num] = true;
37     res.push_back(num);
38     while (cnt < p-1) {
39         int first = 2 * num % p, second = 3 * num % p;
40         if (visit[first] == false && first) {
41             res.push_back(first);
42             visit[first] = true;
43             num = first;
44             cnt++;
45         }
46         else if (visit[second] == false && second) {
47             res.push_back(second);
48             visit[second] = true;
49             num = second;
50             cnt++;
51         } 
52         else {
53             cout << "-1\n";
54             return;
55         }
56     }
57     for (int i = 0; i < res.size(); i++) {
58         printf("%d ", res[i]);
59     }
60     puts(""); 
61 }
62 
63 int main() {
64     int t = read();
65     while (t--) {
66         solve();
67     }
68     return 0;
69 }