[Baltic2014]friends

嘟嘟嘟

首先想想暴力的做法，枚举加入的字符，然后判断删去这个字符后两个长度为n / 2的字符串是否相等，复杂度O(n²)。

所以可以想办法把判断复杂度降低到O(1)，那自然就想到hash了。hash是能做到O(n)预处理，然后O(1)比较的。

取一段的hash值：hash[L, R] = hash[1, R] - hash[1, L - 1] * base^{R - L + 1}.这个也好理解，就是前面的hash[1, L - 1]为整个hash值贡献了hash[1, L - 1] * base^{R - L + 1}，减去即可。

思路就这么明显~~只不过合并hash值得时候得想清楚了……

 #include<cstdio>
 #include<iostream>
 #include<cmath>
 #include<algorithm>
 #include<cstring>
 #include<cstdlib>
 #include<cctype>
 #include<vector>
 #include<stack>
 #include<queue>
 using namespace std;
 #define enter puts("")
 #define space putchar(' ')
 #define Mem(a) memset(a, 0, sizeof(a))
 typedef long long ll;
 typedef unsigned long long ull;
 typedef double db;
 const int INF = 0x3f3f3f3f;
 const ull base = ;
 const int eps = 1e-;
 const int maxn = 2e6  + ;
 inline ll read()
 {
     ll ans = ;
     char ch = getchar(), last = ' ';
     while(!isdigit(ch)) {last = ch; ch = getchar();}
     while(isdigit(ch)) {ans = ans *  + ch - ''; ch = getchar();}
     if(last == '-') ans = -ans;
     return ans;
 }
 inline void write(ll x)
 {
     if(x < ) x = -x, putchar('-');
     if(x >= ) write(x / );
     putchar(x %  + '');
 }

 int n, mid;
 char s[maxn];
 ull has[maxn], f[maxn], las = ;
 int del, tot = ;

 ull get(int L, int R)
 {
     return has[R] - has[L - ] * f[R - L + ];
 }
 int judge(int x)
 {
     ull h1, h2;
     if(x < mid)
     {
         h1 = get(, x - ) * f[mid - x] + get(x + , mid);
         h2 = get(mid + , n);
     }
     else if(x > mid)
     {
         h1 = get(, mid - );
         h2 = get(mid, x - ) * f[n - x] + get(x + , n);
     }
     else
     {
         h1 = get(, x - );
         h2 = get(x + , n);
     }
     if(h1 == h2)
     {
         del = x;
         if(h1 == las) return ;
         las = h1; return ;
     }
     return ;
 }

 int main()
 {
     n = read(); scanf("%s", s + );
     if(!(n & )) {puts("NOT POSSIBLE"); return ;}
     f[] = ;
     for(int i = ; i <= n; ++i) f[i] = f[i - ] * base;
     for(int i = ; i <= n; ++i) has[i] = has[i - ] * base + s[i] - 'A' + ;
     mid = (n >> ) + ;
     for(int i = ; i <= n; ++i) tot += judge(i);
     if(!tot) puts("NOT POSSIBLE");
     else if(tot > ) puts("NOT UNIQUE");
     else
     {
         for(int i = , cnt = ; cnt <= (n >> ); ++i, cnt++)
         {
             if(i == del) cnt--;
             else putchar(s[i]);
         }
         enter;
     }
         return ;
 }