这个范围给的很像区间dp之类的，想了半天没一点思路，滚去看了一眼status被吓傻了。然后瞎猜了一发结论就过掉了。
求出逆序对数，判断是否为奇数即可。因为翻转区间会把将这段区间的逆序对取反，而长度为4x+2和4x+3的区间的数对数量是奇数，所以每次增加或减少的逆序对个数是奇数。
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
#define N 55
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<''||c>'')) c=getchar();return c;}
int gcd(int n,int m){return m==?n:gcd(m,n%m);}
int read()
{
int x=,f=;char c=getchar();
while (c<''||c>'') {if (c=='-') f=-;c=getchar();}
while (c>=''&&c<='') x=(x<<)+(x<<)+(c^),c=getchar();
return x*f;
}
int n,a[N],ans;
int main()
{
#ifndef ONLINE_JUDGE
freopen("bzoj4975.in","r",stdin);
freopen("bzoj4975.out","w",stdout);
const char LL[]="%I64d\n";
#else
const char LL[]="%lld\n";
#endif
n=read();
for (int i=;i<=n;i++) a[i]=read();
for (int i=;i<=n;i++)
for (int j=i+;j<=n;j++)
ans+=a[i]<a[j];
if (ans&) cout<<'Q';else cout<<'T';
return ;
}