Codeforces Round #485 (Div. 1) B. Petr and Permutations_ba82586628365094的博客-程序员秘密

B. Petr and Permutations

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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 $\mathrm{11 to nn and\; then 3n3n times\; takes\; a\; random\; pair\; of\; different\; elements\; and\; swaps\; them.\; Alex\; envies\; Petr\; and\; tries\; to\; imitate\; him\; in\; all\; kind\; of\; things.\; Alex\; has\; also\; come\; up\; with\; a\; problem\; about\; random\; permutation.\; He\; generates\; a\; random\; permutation\; just\; like\; Petr\; but\; swaps\; elements 7n+17n+1 times\; instead\; of 3n3n times.\; Because\; it\; is\; more\; random,\; OK?!}$

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

Input

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 $\mathrm{nn distinct\; integers\; between 11 and nn —\; the\; permutation\; of\; size nn from\; the\; test.}$

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

Output

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).

Example

input

Copy

5
2 4 5 1 3

output

Copy

Petr

Note

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：逆序对的奇偶性等价于交换次数。显然可以树状数组求逆序对个数。看完题解学到一个O(n)的做法，对于序列a，从i向a[i]连条边，在这个n条边的图中统计环的个数。环个数变化的奇偶性就等于逆序对变化的奇偶性。

因为对于每次交换，如果两个点在一个环中，那么环就会被拆分成两个；如果两个点在两个环中，操作后就会变成一个环。代码很简单，直接粘的题解给的代码。

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cmath>
#include <vector>
#include <set>
#include <map>
#include <unordered_set>
#include <unordered_map>
#include <queue>
#include <ctime>
#include <cassert>
#include <complex>
#include <string>
#include <cstring>
using namespace std;

#ifdef LOCAL
    #define eprintf(...) fprintf(stderr, __VA_ARGS__)
#else
    #define eprintf(...) 42
#endif

typedef long long ll;
typedef pair<int, int> pii;
#define mp make_pair

const int N = (int)1e6 + 7;
int n;
int a[N];
int ans;

int main()
{
//    freopen("input.txt", "r", stdin);
//    freopen("output.txt", "w", stdout);

    scanf("%d", &n);
    for (int i = 0; i < n; i++) {
        scanf("%d", &a[i]);
        a[i]--;
    }
    ans = 0;
    for (int i = 0; i < n; i++) {
        if (a[i] == -1) continue;
        ans ^= 1;
        int x = i;
        while(x != -1) {
            int y = a[x];
            a[x] = -1;
            x = y;
        }
    }
    if (ans)
        printf("Um_nik\n");
    else
        printf("Petr\n");

    return 0;
}