哇,你造吗。。。蒟蒻当年NOI这道题。。。可是拿了0分哦~(废话这么多,你弱怪我啊!)

我们先kmp一次,记录下next数组及cnt数组,其中cnt表示他前面可以匹配的模式串个数。

然后在类似kmp的做一次,记录下Next数组,Next == next + 长度限制。

于是。。。ans = Π (cnt[Next[i]] + 1) (1 ≤ i ≤ len)

 

 1 /**************************************************************
 2     Problem: 3670
 3     User: rausen
 4     Language: C++
 5     Result: Accepted
 6     Time:664 ms
 7     Memory:13500 kb
 8 ****************************************************************/
 9  
10 #include <cstdio>
11 #include <cstring>
12  
13 using namespace std;
14 typedef long long ll;
15 const int Len = 1000005;
16 const ll mod = 1000000007;
17  
18 char st[Len];
19 int len, next[Len], Next[Len], cnt[Len];
20 ll ans;
21  
22 inline void work_next() {
23     int i, t;
24     for (i = 1; i <= len; ++i) {
25         t = next[i - 1];
26         while (t != -1 && st[i] != st[t + 1])
27             t = next[t];
28         next[i] = t + 1, cnt[i] = cnt[t + 1] + 1;
29     }
30 }
31  
32 inline void work_Next() {
33     int i, t;
34     for (i = 1; i <= len; ++i) {
35         t = Next[i - 1];
36         if (t * 2 + 2 > i) t = next[t];
37         while (t != -1 && st[i] != st[t + 1])
38             t = next[t];
39         Next[i] = t + 1;
40     }
41 }
42  
43 int main() {
44     int T, i;
45     scanf("%d\n", &T);
46     while (T--) {
47         gets(st + 1), len = strlen(st + 1);
48         next[0] = Next[0] = -1;
49         work_next();
50         work_Next();
51         for (i = ans = 1; i <= len; ++i)
52             (ans *= (cnt[Next[i]] + 1)) %= mod;
53         printf("%lld\n", ans);
54     }
55 }
View Code

 

posted on 2014-12-01 19:35  Xs酱~  阅读(2211)  评论(0编辑  收藏  举报