【HDU-3336-count the string】（KMP）

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
解析：
/*
因爲next數組表示的是子串中最長公共前後綴串的長度，
如果用dt[i]表示該字符串前i個字符中出現任意以第i個字符結尾的前綴的次數，
它的遞推式是 dt[i]=d[next[i]]+1,
即以第i個字符結尾的前綴數等於以第next[i]個字符爲結尾的前綴數加上它自己本身
*/

#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
char str1[200010];
int next1[200010];
int N;
int dt[200010];
void get_next()
{
    int i=0,j=0;
    next1[i]=-1;
    j=next1[i];
    while(i<N)
    {
        if(j==-1||str1[i]==str1[j])
        next1[++i]=++j;
        else j=next1[j];
    }
}
/*
因爲next數組表示的是子串中最長公共前後綴串的長度，
如果用dt[i]表示該字符串前i個字符中出現任意以第i個字符結尾的前綴的次數，
它的遞推式是 dt[i]=d[next[i]]+1,
即以第i個字符結尾的前綴數等於以第next[i]個字符爲結尾的前綴數加上它自己本身
*/
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&N);
        scanf("%s",str1);
        memset(dt,0,sizeof(dt));
        get_next();
        int  sum=0;
        for(int i=1;i<=N;i++)
        {
            dt[i]=dt[next1[i]]+1;
            sum=(sum+dt[i])%10007;
         } 
         printf("%d\n",sum);
    }
    return 0;
}