Petr likes to come up with problems about randomly generated data. This time problem is about random permutation. He decided to generate a random permutation this way: he takes identity permutation of numbers from

You somehow get a test from one of these problems and now you want to know from which one.

In the first line of input there is one integer $\mathrm{nn\; (103\le n\le 106103\le n\le 106).}$

In the second line there are $\mathrm{nn\; distinct\; integers\; between\; 11\; and\; nn\; —\; the\; permutation\; of\; size}$

It is guaranteed that all tests except for sample are generated this way: First we choose $\mathrm{nn\; —\; the\; size\; of\; the\; permutation.\; Then\; we\; randomly\; choose\; a\; method\; to\; generate\; a\; permutation\; —\; the\; one\; of\; Petr\; or\; the\; one\; of\; Alex.\; Then\; we\; generate\; a\; permutation\; using\; chosen\; method.}$

If the test is generated via Petr‘s method print "Petr" (without quotes). If the test is generated via Alex‘s method print "Um_nik" (without quotes).

5
2 4 5 1 3

Petr

Please note that the sample is not a valid test (because of limitations for $\mathrm{nn)\; and\; is\; given\; only\; to\; illustrate\; input/output\; format.\; Your\; program\; still\; has\; to\; print\; correct\; answer\; to\; this\; test\; to\; get\; AC.}$

Due to randomness of input hacks in this problem are forbidden.

思路：逆序對的奇偶性等價於交換次數。顯然可以樹狀數組求逆序對個數。看完題解學到一個O(n)的做法，對於序列a，從i向a[i]連條邊，在這個n條邊的圖中統計環的個數。環個數變化的奇偶性就等於逆序對變化的奇偶性。

因為對於每次交換，如果兩個點在一個環中，那麽環就會被拆分成兩個；如果兩個點在兩個環中，操作後就會變成一個環。代碼很簡單，直接粘的題解給的代碼。

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstdlib>
 4 #include <algorithm>
 5 #include <cmath>
 6 #include <vector>
 7 #include <set>
 8 #include <map>
 9 #include <unordered_set>
10 #include <unordered_map>
11 #include <queue>
12 #include <ctime>
13 #include <cassert>
14 #include <complex>
15 #include <string>
16 #include <cstring>
17 using namespace std;
18 
19 #ifdef LOCAL
20     #define eprintf(...) fprintf(stderr, __VA_ARGS__)
21 #else
22     #define eprintf(...) 42
23 #endif
24 
25 typedef long long ll;
26 typedef pair<int, int> pii;
27 #define mp make_pair
28 
29 const int N = (int)1e6 + 7;
30 int n;
31 int a[N];
32 int ans;
33 
34 int main()
35 {
36 //    freopen("input.txt", "r", stdin);
37 //    freopen("output.txt", "w", stdout);
38 
39     scanf("%d", &n);
40     for (int i = 0; i < n; i++) {
41         scanf("%d", &a[i]);
42         a[i]--;
43     }
44     ans = 0;
45     for (int i = 0; i < n; i++) {
46         if (a[i] == -1) continue;
47         ans ^= 1;
48         int x = i;
49         while(x != -1) {
50             int y = a[x];
51             a[x] = -1;
52             x = y;
53         }
54     }
55     if (ans)
56         printf("Um_nik\n");
57     else
58         printf("Petr\n");
59 
60     return 0;
61 }

Codeforces Round #485 (Div. 1) B. Petr and Permutations