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#include#include#include#include#include#include#include#include#include#include#include#include#include#include#include#include//#include//#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-4typedef pair pii;#define pi acos(-1.0)//const int N = 1005;#define REP(i,n) for(int i=0;i<(n);i++)typedef pair 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;} EPFL - Fighting
