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Codeforces Round #485 (Div. 1) B. Petr and Permutations 思维
阅读量:614 次
发布时间:2019-03-13

本文共 1720 字,大约阅读时间需要 5 分钟。

// Siberian Squirrel#include 
#define IO ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)#define ACM_LOCAL using namespace std;typedef long long ll;const int INF = 0x3f3f3f3f;const int N = 2e6 + 10;const int MOD = 998244353;ll n, m, k, w, res = 0;int f[N], a[N], q[N], t[N];ll quick_pow(ll ans, ll p, ll res = 1) { for(; p; p >>= 1, ans = ans * ans % MOD) if(p & 1) res = res * ans % MOD; return res % MOD;}ll inv(ll ans) { return quick_pow(ans, MOD - 2);}ll C(int n, int m) { if(n < m || n <= 0 || m < 0) return 0; if(m == 0) return 1; return (ll)f[n] * inv(f[m]) % MOD * inv(f[n - m]) % MOD;}void merge_sort(int l, int r) { if(l >= r) return; int mid = l + r >> 1; merge_sort(l, mid), merge_sort(mid + 1, r); int k = 0, i = l, j = mid + 1; while(i <= mid && j <= r) { if(q[i] <= q[j]) { t[k++] = q[i++]; } else { t[k++] = q[j++]; res += mid - i + 1; } } while(i <= mid) t[k++] = q[i++]; while(j <= r) t[k++] = q[j++]; for(i = l, j = 0; i <= r; ++i, ++j) q[i] = t[j];}inline void solve(ll ans = 0) { if((abs(((n % 2) - (res % 2)))) & 1) { cout << "Um_nik\n"; } else { cout << "Petr\n"; }}int main() { IO; #ifdef ACM_LOCAL freopen("input.txt", "r", stdin); freopen("output.txt", "w", stdout); #endif f[0] = f[1] = 1; for(int i = 2; i < N; ++i) { f[i] = (ll)f[i - 1] * i % MOD; } int o = 1; while(o--) { cin >> n; for(int i = 0; i < n; ++i) { cin >> q[i]; } merge_sort(0, n - 1); solve(); } return 0;}

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