Count the string（KMP）

Count the string
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 11064 Accepted Submission(s): 5156

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

1
4
abab

Sample Output

6

#include<iostream>#include<string>#include<stdio.h>#include<string.h>using namespace std;int nxt[200007],lenp;char p[200007];int dt[200007],sum;void getNext(){    int i =0,j =-1;    nxt[0] = -1;    while(i<lenp)    {        if(j==-1||p[i]==p[j])            nxt[++i] =++j;        else            j = nxt[j];    }}void kmp(){    int i =0;    memset(dt,0,sizeof(dt));    sum = 0;    //dt数组的意义是统计以i结尾的前缀出现的次数    //而nxt数组存储的就是最大前后缀的长度    //所以dt[i] = dt[nxt[i]] +1 意思就是第i位的次数 就是dt[nxt[i]] + 本身    //举个例子 aba 以a结尾前缀的有 a aba 这两种 所以dt[3] = 2 ababa 以a为结尾前缀的有a aba ababa 这三种 dt[5] =3;    for(int i =1;i<=lenp;i++)    {        dt[i] = dt[nxt[i]] +1;        sum = (sum + dt[i])%10007;    }}int main(){    int ncase;    scanf("%d",&ncase);    while(ncase--)    {        scanf("%d",&lenp);        scanf("%s",p);        getNext();        kmp();        printf("%d\n",sum);    }    return 0;}