이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include <bits/stdc++.h>
#define FOR(i, a, b) for (int i = (a); i < (b); i++)
#define FORd(i, a, b) for (int i = (a); i >= (b); i--)
#define REP(i, n) FOR(i, 0, n)
#define ll long long
using namespace std;
const int MAXN = 2001000;
ll ha[MAXN], pot[MAXN];
ll ha2[MAXN], pot2[MAXN];
const int BASE = 35537;
const int MOD = 1e9 + 7;
const int BASE2 = 57737;
const int MOD2 = 1e9 + 9;
ll val(int x, int y) {
if (y < x) return 0;
ll out = ha[y];
if (x) out = (((out - ha[x - 1] * pot[y - x + 1]) % MOD) + MOD) % MOD;
return out;
}
ll val2(int x, int y) {
if (y < x) return 0;
ll out = ha2[y];
if (x) out = (((out - ha2[x - 1] * pot2[y - x + 1]) % MOD2) + MOD2) % MOD2;
return out;
}
int main() {
ios_base::sync_with_stdio(false);
pot[0] = 1;
FOR(i, 1, MAXN) pot[i] = (pot[i - 1] * BASE) % MOD;
pot2[0] = 1;
FOR(i, 1, MAXN) pot2[i] = (pot2[i - 1] * BASE2) % MOD2;
int n;
cin >> n;
string s;
cin >> s;
ll curr = 0, curr2 = 0;
REP(i, n) {
curr = (curr * BASE + s[i]) % MOD;
ha[i] = curr;
curr2 = (curr2 * BASE2 + s[i]) % MOD2;
ha2[i] = curr2;
}
string out = "";
ll V = -1, V2 = -1;
REP(i, n) {
if (i <= n / 2) {
ll a = (val(0, i - 1) * pot[n / 2 - i] + val(i + 1, n / 2)) % MOD;
ll b = val(n / 2 + 1, n - 1);
ll a2 = (val2(0, i - 1) * pot2[n / 2 - i] + val2(i + 1, n / 2)) % MOD2;
ll b2 = val2(n / 2 + 1, n - 1);
if (a != b) continue;
if (a2 != b2) continue;
string s1 = "", s2 = "";
REP(x, i) s1 += s[x];
FOR(x, i + 1, n / 2 + 1) s1 += s[x];
FOR(x, n / 2 + 1, n) s2 += s[x];
assert(s1 == s2);
if (out != "" && (V != a || V2 != a2)) {
cout << "NOT UNIQUE\n";
return 0;
} else if (out == "") {
V = a;
V2 = a2;
out = s.substr(n / 2 + 1, n / 2);
}
} else {
ll a = val(0, n / 2 - 1);
ll b = (val(n / 2, i - 1) * pot[n - i - 1] + val(i + 1, n - 1)) % MOD;
ll a2 = val2(0, n / 2 - 1);
ll b2 = (val2(n / 2, i - 1) * pot2[n - i - 1] + val2(i + 1, n - 1)) % MOD2;
if (a != b) continue;
if (a2 != b2) continue;
string s1 = "", s2 = "";
REP(x, n / 2) s1 += s[x];
FOR(x, n / 2, i) s2 += s[x];
FOR(x, i + 1, n) s2 += s[x];
assert(s1 == s2);
if (out != "" && (V != a || V2 != a2)) {
cout << "NOT UNIQUE\n";
return 0;
} else if (out == "") {
V = a;
V2 = a2;
out = s.substr(0, n / 2 - 1);
}
}
}
if (out == "") cout << "NOT POSSIBLE\n";
else cout << out << "\n";
return 0;
}
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