CF986B Petr and Permutations 思维

每次交换：逆序对的数量+1或者-1；

假设最后逆序对数量为 sum;

①x+y=3n;

②x-y=sum;

-> 3n+sum为偶数；

所以 n 和 sum 必须奇偶一样；

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<bitset>
#include<ctime>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
//#include<cctype>
//#pragma GCC optimize(2)
using namespace std;
#define maxn 1000005
#define inf 0x7fffffff
//#define INF 1e18
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
#define rdult(x) scanf("%lu",&x)
#define rdlf(x) scanf("%lf",&x)
#define rdstr(x) scanf("%s",x)
typedef long long  ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const long long int mod = 1e9 + 7;
#define Mod 1000000000
#define sq(x) (x)*(x)
#define eps 1e-4
typedef pair<int, int> pii;
#define pi acos(-1.0)
//const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)
typedef pair<int, int> pii;
inline ll rd() {
    ll x = 0;
    char c = getchar();
    bool f = false;
    while (!isdigit(c)) {
        if (c == '-') f = true;
        c = getchar();
    }
    while (isdigit(c)) {
        x = (x << 1) + (x << 3) + (c ^ 48);
        c = getchar();
    }
    return f ? -x : x;
}

ll gcd(ll a, ll b) {
    return b == 0 ? a : gcd(b, a%b);
}
int sqr(int x) { return x * x; }

/*ll ans;
ll exgcd(ll a, ll b, ll &x, ll &y) {
    if (!b) {
        x = 1; y = 0; return a;
    }
    ans = exgcd(b, a%b, x, y);
    ll t = x; x = y; y = t - a / b * y;
    return ans;
}
*/

int n;
int a[maxn];
int b[maxn];
int c[maxn];

void add(int x) {
    while (x <= n) {
        c[x]++; x += x & -x;
    }
}
int query(int x) {
    int res = 0;
    while (x > 0) {
        res += c[x]; x -= x & -x;
    }
    return res;
}
int main() {
    //ios::sync_with_stdio(0);
    rdint(n);
    for (int i = 1; i <= n; i++) {
        rdint(a[i]); b[i] = a[i];
    }
    sort(b + 1, b + 1 + n);
    int ans = 0;
    for (int i = 1; i <= n; i++) {
        add(lower_bound(b + 1, b + 1 + n, a[i]) - b);
        ans += (i - query(lower_bound(b + 1, b + 1 + n, a[i] + 1) - b - 1));
    }
    if ((ans & 1) == (n & 1)) {
        cout << "Petr" << endl;
    }
    else cout << "Um_nik" << endl;
    return 0;
}