Codeforces Round #485 (Div. 2) E. Petr and Permutations
题目连接：
http://codeforces.com/contest/987/problem/E
Description
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 $1$ to $n$ and then $3n$ 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 $7n+1$ times instead of $3n$ 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.
Sample Input
5
2 4 5 1 3Sample Output
Petr题意
给定一个排列，问它是从\(1, 2, 3,...,n\)变换\(3n\)次而来，还是\(7n+1\)
Give a random permutation, it maybe transform \(3n\) times or \(7n+1\) times from the permutation of numbers from \(1\) to n. Judging it is \(3n\) or \(7n+1\) ?
题解：
考虑n是奇数的情况，\(3n\)为奇数，\(7n+1\)为偶数；n是偶数时，\(3n\)为偶数，\(7n+1\)为奇数。
考虑最小交换次数，假定为k，\(k+2*t\)次交换都可以变成目标串，但奇偶性相同。所以只需判断最小交换次数的奇偶性即可。
When n is a odd number, \(3n\) is odd，\(7n+1\) is even.When n is a even number, \(3n\) is even，\(7n+1\) is odd.
Consider about the swap times, assume k is the mininum swap times to the final permutation , \(k+2*t\) will reach the final result , \(k\) and \(k+2*t\) have the same parity. So we need to know the parity of the mininum swap times.
如果我们不把stdio的同步关闭，将会读入超时
If we dont close the sync with stdio , time limit exceeded.
代码
#include <bits/stdc++.h>
using namespace std;
int n;
map<int,int> m;
int a[1000010];
int c[1000010];
int ans;
inline int lowbit(int k) {
return (k&(-k));
}
void update(int t) {
while (t<n) {
c[t]++;
t+=lowbit(t);
}
}
int sum(int t) {
int ret=0;
while (t>0) {
ret+=c[t];
t-=lowbit(t);
}
return ret;
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(nullptr);
cout.tie(nullptr);
cerr.tie(nullptr);
cin>>n;
for (int i=1;i<=n;i++) {
cin>>a[i];
m[a[i]]=i;
}
for (int i=n;i;i--) {
ans += i-(m[i]-sum(m[i]-1));
update(m[i]);
}
if ((n&1)^(ans&1)==0) cout << "Petr" <<endl;
else cout << "Um_nik" <<endl;
}原文地址：https://www.cnblogs.com/EDGsheryl/p/9157964.html