hdu 3336 Count the string

http://acm.hdu.edu.cn/showproblem.php?pid=3336

Count the string

Problem Description

It is well known that AekdyCoin is good at string problems as well as number theory problems. When given a string s, we can write down all the non-empty prefixes of this string. For example:
s: "abab"
The prefixes are: "a", "ab", "aba", "abab"
For each prefix, we can count the times it matches in s. So we can see that prefix "a" matches twice, "ab" matches twice too, "aba" matches once, and "abab" matches once. Now you are asked to calculate the sum of the match times for all the prefixes. For "abab", it is 2 + 2 + 1 + 1 = 6.
The answer may be very large, so output the answer mod 10007.

Input

The first line is a single integer T, indicating the number of test cases.
For each case, the first line is an integer n (1 <= n <= 200000), which is the length of string s. A line follows giving the string s. The characters in the strings are all lower-case letters.

Output

For each case, output only one number: the sum of the match times for all the prefixes of s mod 10007.

Sample Input

14abab

Sample Output

6

如果用dt[i]表示该字符串前i个字符中出现任意以第i个字符结尾的前缀的次数，它的递推式是 dt[i]=d[next[i]]+1,即以第i个字符结尾的前缀数等于以第next[i]个字符为结尾的前缀数加上它自己本身

举个例子：
           i 1  2  3  4  5  6
      字符串 a  b  a  b  a  b
       dt[i] 1  1  2  2  3  3
        aba中出现的前缀为a,aba,所以num[3]是2，ababa中出现的前缀为a,aba,ababa,所以num[5]是3，当i=5时,Next[5]=3，所以num[i]=num[Next[i]]+1
理解了上面的部分就很简单了，后面直接套next数组的模板

#include <iostream>#include <cstring>#include <cstdio>#include <algorithm>#include <cmath>#include <cstdlib>#include <limits>#include <queue>#include <stack>#include <vector>#include <map>using namespace std;#define N 254010#define INF 0x3f3f3f3f#define PI acos (-1.0)#define EPS 1e-8#define met(a, b) memset (a, b, sizeof (a))char s[N];int Next[N], num[N], len;void get_next ();int main (){    int t;    scanf ("%d", &t);    while (t--)    {        int sum = 0;        scanf ("%d %s", &len, s);        get_next ();        for (int i=1; i<=len; i++)        {            num[i] = (num[ Next[i] ] + 1) % 10007;            sum += num[i];            sum %= 10007;        }        printf ("%d\n", sum);    }    return 0;}void get_next (){    int i = 0, j = -1;    Next[i] = -1;    while (i < len)    {        if (j == -1 || s[i] == s[j])            Next[++i] = ++j;        else j = Next[j];    }    return;}