이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include<iostream>
#include<iomanip>
#include<algorithm>
#include<stack>
#include<queue>
#include<string>
#include<string.h>
#include<cmath>
#include<vector>
#include<map>
#include<unordered_map>
#include<set>
#include<unordered_set>
#include<cstdio>
#include<bitset>
#include<chrono>
#include<random>
#include<ext/rope>
/* ordered_set
#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
#define ordered_set tree<int, null_type,less<int>, rb_tree_tag,tree_order_statistics_node_update>
*/
#define pb push_back
#define fi first
#define se second
using namespace std;
using ll = long long;
using ld = long double;
using ull = unsigned long long;
const int maxN = 3e6 + 5;
const int mod = 1e9 + 7;
const ll oo = 1e18;
#define int long long
string s;
const int base = 31;
int n;
ll Hash[maxN], poww[maxN];
inline void FastInput()
{
ios_base::sync_with_stdio(false);
cin.tie(nullptr);
}
void Prepair()
{
poww[0] = 1;
for(int i=1; i<maxN; i++) poww[i] = (poww[i - 1] * base) % mod;
}
void ReadInput()
{
cin >> n >> s;
s = ' ' + s;
for(int i=1; i<=n; i++)
Hash[i] = (Hash[i - 1] * base + s[i] - 'a' + 1) % mod;
}
int gethash(int l, int r)
{
return (Hash[r] - Hash[l - 1] * poww[r - l + 1] + 1LL * mod * mod) % mod;
}
ll Power(ll a, ll b)
{
if(b == 0) return 1;
ll t = (Power(a, b / 2));
if(b & 1) return t * t % mod * a % mod;
return t * t % mod;
}
void Solve()
{
if(!(n & 1))
{
cout << "NOT POSSIBLE";
return;
}
vector<int> res;
for(int i=1; i<=n; i++)
{
ll _hash1, _hash2;
if(i <= n / 2)
{
ll t1 = gethash(1, i - 1);
ll t2 = gethash(i + 1, n / 2 + 1);
t1 *= poww[n / 2 - i + 1];
t1 %= mod;
_hash1 = (t1 + t2) % mod;
_hash2 = gethash(n / 2 + 2, n);
}
else
{
_hash1 = gethash(1, n / 2);
ll t1 = gethash(n / 2 + 1, i - 1);
ll t2 = gethash(i + 1, n);
t1 *= poww[n - i];
t1 %= mod;
_hash2 = (t1 + t2) % mod;
}
if(_hash1 == _hash2) res.pb(i);
}
if(res.size() > 1)
{
string s1, s2;
s1 = s;
s2 = s;
s1.erase(s1.begin() + 1);
s2.pop_back();
if(res[0] == 1 && res.back() == n && s1 != s2)
{
cout << "NOT UNIQUE";
return;
}
}
if(res.empty())
{
cout << "NOT POSSIBLE";
return;
}
s.erase(s.begin() + res[0]);
for(int i=1; i<=n/2; i++) cout << s[i];
}
int32_t main()
{
//freopen("x.inp", "r", stdin);
FastInput();
Prepair();
ReadInput();
Solve();
}
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