P2375 [NOI2014] 动物园
如果没有“不重叠”的限制，即为 KMP 的 next 数组。现在要求长度不能超过串的一半，那么就反复跳 next 直到得到答案为止。
查看代码
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
| #include <cstdio>
#include <cstring>
using namespace std;
template <class Type>
void read (Type &x)
{
char c;
bool flag = false;
while ((c = getchar()) < '0' || c > '9')
if (c == '-') flag = true;
x = c - '0';
while ((c = getchar()) >= '0' && c <= '9')
x = (x << 3) + (x << 1) + c - '0';
if (flag) x = ~x + 1;
}
template <class Type, class ...Rest>
void read (Type &x, Rest &...y) { read(x), read(y...); }
template <class Type>
void write (Type x)
{
if (x < 0) putchar('-'), x = ~x + 1;
if (x > 9) write(x / 10);
putchar('0' + x % 10);
}
typedef long long L;
const int N = 1e6 + 10, mod = 1e9 + 7;
char s[N];
int n, nxt[N], f[N];
int main ()
{
int T; read(T);
while (T--)
{
scanf("%s", s + 1);
n = strlen(s + 1);
f[1] = 1;
for (int i = 2, j = 0; i <= n; ++i)
{
while (j && s[i] != s[j + 1]) j = nxt[j];
if (j < n && s[i] == s[j + 1]) ++j;
nxt[i] = j;
f[i] = f[j] + 1;
}
int res = 1;
for (int i = 2, j = 0; i <= n; ++i)
{
while (j && s[i] != s[j + 1]) j = nxt[j];
if (j < n && s[i] == s[j + 1]) ++j;
while (j << 1 > i) j = nxt[j];
res = (L)res * (f[j] + 1) % mod;
}
write(res), puts("");
}
return 0;
}
|