BZOJ 3916: [Baltic2014]friends( hash )

字符串哈希..然后枚举每一位+各种判断就行了

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#include<cstdio>#include<cstring>#include<algorithm> using namespace std; typedef long long ll; const int MOD = 1000000007;const int P = 59;const int maxn = 2000009; int N, H[maxn], Power[maxn], n = 0;char s[maxn]; int getH(int p, int len) { return (H[p] - ll(H[p + len]) * Power[len] % MOD + MOD) % MOD;} int main() { scanf("%d%s", &N, s); H
= 0; for(int i = N; i--; ) H[i] = (ll(P) * H[i + 1] + s[i]) % MOD; Power[0] = 1; for(int i = 1; i <= N; i++) Power[i] = ll(P) * Power[i - 1] % MOD; if(N % 2 == 0) { puts("NOT POSSIBLE"); return 0; } int ans0 = -1, ans1 = -1, h = H[N / 2 + 1]; for(int i = N >> 1; i--; ) if((ll(Power[i]) * getH(i + 1, N / 2 - i) % MOD + getH(0, i)) % MOD == h) { ans0 = i; break; } h = getH(0, N >> 1); for(int i = N >> 1; ++i < N; ) if((ll(Power[i - N / 2]) * getH(i + 1, N - i - 1) % MOD + getH(N / 2, i - N / 2)) % MOD == h) { ans1 = i; break; } if(~ans0 && ~ans1) { if(getH(0, N >> 1) != H[N / 2 + 1]) puts("NOT UNIQUE"); else { for(int i = 0; i < (N / 2); i++) putchar(s[i]); } } else if(!~ans0 && !~ans1) { if(getH(0, N >> 1) == H[N / 2 + 1]) { for(int i = 0; i < (N / 2); i++) putchar(s[i]); } else { puts("NOT POSSIBLE"); } } else if(~ans0) { for(int i = N >> 1; ++i < N; ) putchar(s[i]); } else { for(int i = 0; i < N / 2; i++) putchar(s[i]); } return 0;}--------------------------------------------------------------------------------

3916: [Baltic2014]friends

Time Limit: 30 Sec Memory Limit: 256 MB
Submit: 403 Solved: 148
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Description

有三个好朋友喜欢在一起玩游戏,A君写下一个字符串S,B君将其复制一遍得到T,C君在T的任意位置(包括首尾)插入一个字符得到U.现在你得到了U,请你找出S.

Input

第一行一个数N,表示U的长度.第二行一个字符串U,保证U由大写字母组成

Output

输出一行,若S不存在,输出"NOT POSSIBLE".若S不唯一,输出"NOT UNIQUE".否则输出S.

Sample Input

Sample Input1:
7
ABXCABC

Sample Input2:

6
ABCDEF

Sample Input3:

9
ABABABABA

Sample Output

Sample Output1:

ABC

Sample Output2:

NOT POSSIBLE

Sample Output3:

NOT UNIQUE

HINT

对于100%的数据 2<=N<=2000001

Source