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] * baseR - 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] * baseR - 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……
1 #include<cstdio> 2 #include<iostream> 3 #include<cmath> 4 #include<algorithm> 5 #include<cstring> 6 #include<cstdlib> 7 #include<cctype> 8 #include<vector> 9 #include<stack> 10 #include<queue> 11 using namespace std; 12 #define enter puts("") 13 #define space putchar(' ') 14 #define Mem(a) memset(a, 0, sizeof(a)) 15 typedef long long ll; 16 typedef unsigned long long ull; 17 typedef double db; 18 const int INF = 0x3f3f3f3f; 19 const ull base = 19260817; 20 const int eps = 1e-8; 21 const int maxn = 2e6 + 5; 22 inline ll read() 23 { 24 ll ans = 0; 25 char ch = getchar(), last = ' '; 26 while(!isdigit(ch)) {last = ch; ch = getchar();} 27 while(isdigit(ch)) {ans = ans * 10 + ch - '0'; ch = getchar();} 28 if(last == '-') ans = -ans; 29 return ans; 30 } 31 inline void write(ll x) 32 { 33 if(x < 0) x = -x, putchar('-'); 34 if(x >= 10) write(x / 10); 35 putchar(x % 10 + '0'); 36 } 37 38 int n, mid; 39 char s[maxn]; 40 ull has[maxn], f[maxn], las = 0; 41 int del, tot = 0; 42 43 ull get(int L, int R) 44 { 45 return has[R] - has[L - 1] * f[R - L + 1]; 46 } 47 int judge(int x) 48 { 49 ull h1, h2; 50 if(x < mid) 51 { 52 h1 = get(1, x - 1) * f[mid - x] + get(x + 1, mid); 53 h2 = get(mid + 1, n); 54 } 55 else if(x > mid) 56 { 57 h1 = get(1, mid - 1); 58 h2 = get(mid, x - 1) * f[n - x] + get(x + 1, n); 59 } 60 else 61 { 62 h1 = get(1, x - 1); 63 h2 = get(x + 1, n); 64 } 65 if(h1 == h2) 66 { 67 del = x; 68 if(h1 == las) return 0; 69 las = h1; return 1; 70 } 71 return 0; 72 } 73 74 int main() 75 { 76 n = read(); scanf("%s", s + 1); 77 if(!(n & 1)) {puts("NOT POSSIBLE"); return 0;} 78 f[0] = 1; 79 for(int i = 1; i <= n; ++i) f[i] = f[i - 1] * base; 80 for(int i = 1; i <= n; ++i) has[i] = has[i - 1] * base + s[i] - 'A' + 1; 81 mid = (n >> 1) + 1; 82 for(int i = 1; i <= n; ++i) tot += judge(i); 83 if(!tot) puts("NOT POSSIBLE"); 84 else if(tot > 1) puts("NOT UNIQUE"); 85 else 86 { 87 for(int i = 1, cnt = 1; cnt <= (n >> 1); ++i, cnt++) 88 { 89 if(i == del) cnt--; 90 else putchar(s[i]); 91 } 92 enter; 93 } 94 return 0; 95 }