https://codeforces.com/problemset/problem/986/B
题目描述
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 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.
输入格式
In the first line of input there is one integer nn ( 10^{3} \le n \le 10^{6}103≤n≤106 ).
In the second line there are 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 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.
输出格式
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).
题意翻译
Petr要打乱排列。他首先有一个从 11 到 nn 的顺序排列，然后进行 3n3n 次操作，每次选两个数并交换它们。
Alex也要打乱排列。他与Petr唯一的不同是他进行 7n+17n+1 次操作。
给定一个 11 到 nn 的排列。问是由谁打乱的。如果是Petr，输出"Petr"，否则输出"Um_nik"（不是Alex）
感谢@AKEE 提供翻译
输入输出样例
输入 #1复制
5 2 4 5 1 3
输出 #1复制
Petr
说明/提示
Please note that the sample is not a valid test (because of limitations for 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.
开始看到题目一脸懵逼.毫无思路。但是看到3*n和7*n+1这么特殊的两个数感觉会考奇偶性..
思路：交换两个数会导致逆序对个数+1/-1.设+1的个数为x，-1的个数为y.
设k为逆序对数量
那么x+y=3*n.x-y=k(这里y-x=k得出的结论是一样的)
2*x=3*n+k.也就是3*n+k是偶数，那么3*n和k的奇偶性是一致的。
求一下i<j&&a[i]>a[j]的k逆序对个数再和3*n判断一下奇偶性一样不一样就好了。
#include<iostream>
#include<vector>
#include<queue>
#include<cstring>
#include<cmath>
#include<map>
#include<set>
#include<cstdio>
#include<algorithm>
#define lowbit(x) x&(-x);
#define debug(a) cout<<#a<<"="<<a<<endl;
using namespace std;
const int maxn=1e6+100;
typedef long long LL;
LL tree[maxn],a[maxn],n;
void add(LL x,LL d)
{
while(x<=n)
{
tree[x]+=d;
x+=lowbit(x);
}
}
LL getsum(LL x)
{
LL sum=0;
while(x>0)
{
sum+=tree[x];
x-=lowbit(x);
}
return sum;
}
int main(void)
{
cin.tie(0);std::ios::sync_with_stdio(false);
cin>>n;
for(LL i=1;i<=n;i++) cin>>a[i];
LL ans=0;
for(LL i=1;i<=n;i++){
add(a[i],1);
ans+=i-getsum(a[i]);
}
// debug(ans);
if( ( (3*n)%2 )==(ans%2) ){
cout<<"Petr"<<endl;
}
else{
cout<<"Um_nik"<<endl;
}
return 0;
}