[Baltic2004]Friends 题解

题意：对于一个字符串(s),复制一遍之后得到(e)，在(e)的任何位置插入一个字符形成(u)。给定(u)，求字符串(s).

字符串哈希。
考虑尝试每一个字符，尝试一下去除这个字符之后剩下的字符串是否能成功分成两部分。
用字符串哈希自带的“拼凑”功能即可。

#pragma GCC optimize("2")
#include <bits/stdc++.h>
#define PI 3.14159265
#define PQ priority_queue
#define lowbit(x) ((x) & (-x))
#define rep(i , x , y)  for(int (i) = (x) ; (i) <= (y) ; ++i)
#define per(i , x , y)  for(int (i) = (x) ; (i) >= (y) ; --i)
//#define lmy
#ifdef lmy
	#define DEBUG printf("Passing [%s] in LINE %d...
" , __FUNCTION__ , __LINE__)
#endif
typedef long long ll;
typedef unsigned long long ull;

using namespace std;

inline void read(int &x) {
	int f = 1;x = 0;char ch = getchar();
	while(ch < '0' || ch > '9') {if(ch == '-') f = -1; ch = getchar();}
	while(ch >= '0' && ch <= '9') {x = x * 10 + ch - '0'; ch = getchar();}
	x *= f;
}

inline void readll(ll &x) {
	ll f = 1;x = 0;char ch = getchar();
	while(ch < '0' || ch > '9') {if(ch == '-') f = -1; ch = getchar();}
	while(ch >= '0' && ch <= '9') {x = x * 10 + ch - '0'; ch = getchar();}
	x *= f;
}

inline void print(int x) {
    if (x < 0) putchar('-'), x = -x;if (x > 9) print(x / 10);putchar(x % 10 + 48);
}
inline void printll(ll x) {
	if (x < 0) putchar('-'), x = -x;if (x > 9) print(x / 10);putchar(x % 10 + 48);
}

const int MaxN = 2000000 + 5;
const ull bse = 13331;
char a[MaxN];
int n;
ull Pw[MaxN] , Hsh[MaxN];
vector <int> p;

void HshPre() {
	Pw[0] = 1;
	rep(i , 1 , n + 1) Pw[i] = Pw[i - 1] * bse;
	rep(i , 1 , n) {
		Hsh[i] = Hsh[i - 1] * bse + (ull)(a[i]);
	}
}

ull HshValue(int l , int r) {
	return Hsh[r] - Hsh[l - 1] * Pw[r - l + 1];
}

ull Merge(int l , int r , int ql , int qr) {
	// BaoZheng l , r < ql , qr
	ull HshValue1 = HshValue(l , r) , HshValue2 = HshValue(ql , qr);
	return HshValue1 * Pw[qr - ql + 1] + HshValue2;
}

bool Chksame(int l , int r , int ql , int qr) {
	return HshValue(l , r) == HshValue(ql , qr);
}

set <ull> obtain;
void solve() {
	cin >> n;
	scanf("%s" , (a + 1));
	//n = strlen(a + 1);
	if(!(n % 2)) {
		puts("NOT POSSIBLE");
		return ;
	}

	HshPre();
	rep(i , 1 , n) {
		if(i == 1) {
			if(Chksame(2 , (n + 1) / 2 , (n + 1) / 2 + 1 , n) && !obtain . count(HshValue(2 , (n + 1) / 2))) {
			//	DEBUG;
				p . push_back(i);
				obtain . insert(HshValue(2 , (n + 1) / 2));
			}
		}
		else if(i == n) {
			if(Chksame(1 , n / 2 , n / 2 + 1 , n - 1) && !obtain . count(HshValue(1 , n / 2))) {
				//DEBUG;
				p . push_back(i);
				obtain . insert(HshValue(1 , n / 2));
			}
		}
		else if(i == (n + 1) / 2) {
			if(Chksame(1 , n / 2 , n - (n / 2) + 1 , n) && !obtain . count(HshValue(1 , n / 2))) {
				p . push_back(i) , obtain . insert(HshValue(1 , n / 2));
			}
		}
		else if(i > (n + 1) / 2) {
			int l = 1 , r = n / 2;
			int pl = r + 1 , pr = i - 1;
			int ppl = i + 1 , ppr = n;
			ull V = Merge(pl , pr , ppl , ppr);
			if(V == HshValue(l , r) && !obtain . count(HshValue(l , r))) {
				p . push_back(i);
				obtain . insert(HshValue(l , r));
			}
		}
		else if(i < (n + 1) / 2) {
			int l = n - (n / 2) + 1 , r = n;
			int pl = 1 , pr = i - 1;
			int ppl = i + 1 , ppr = l - 1;
			ull V = Merge(pl , pr , ppl , ppr);
			//if(i == 3) cout << V << "DBG , " << HshValue(l , r) << " Done
";
			if(V == HshValue(l , r) && !obtain . count(HshValue(l , r))) {
				p . push_back(i);
				obtain . insert(HshValue(l , r));
			}
		}
	}

	if(p . size() == 0) {
		puts("NOT POSSIBLE");
		return ;
	}
	else if(p . size() == 1) {
		int _ = p[0];
		if(_ == 1) {
			rep(i , (n + 1) / 2 + 1 , n) putchar(a[i]);
			return ;
		}
		else if(_ == n) {
			rep(i , 1 , n / 2) putchar(a[i]);
			return ;
		}
		else if(_ == (n + 1) / 2) {
			rep(i , 1 , n / 2) putchar(a[i]);
			return ;
		}
		else if(_ > (n + 1) / 2) {
			rep(i , 1 , n / 2) putchar(a[i]);
			return ;
		}

		else if(_ < (n + 1) / 2) {
			rep(i , (n + 1) / 2 + 1 , n) putchar(a[i]);
			return ;
		}
		return ;
	}
	else if(p . size() > 1) {
		puts("NOT UNIQUE");
		return ;
	}
}

int main(){
	#ifdef lmy
		int nol_cl = clock();
	#endif

	solve();

	// did you forget to undefine Lmy ?
	#ifdef lmy
		printf("
Time: %dms
", int((clock() - nol_cl) / (double)CLOCKS_PER_SEC * 1000));
	#endif
	return 0;
}