K

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. 

InputThe 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. 
OutputFor 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

一个结论：cnt[i] 表示 因为加入i字符而增加的 以i字符结束的前缀数目
cnt[i] = cnt[Next[i]] + 1
Next[]数组表示前j个字符中最长的相同前缀后缀长度
证明：
如果Next[i]==0 也就是说 i字符没有在之前出现过，那么因为加入字符i增加的前缀只有0-i这个字符串
否则  增加的数目要考虑由于第i-k个字符到第i个字符组成的字符串是前缀的情况，增加的数目为这种前缀的数目+1，那么这种前缀的数目如何求？
显然，考虑i之前最长匹配的前缀后缀，在最长匹配处加入字符i增加的前缀 等于 在i处加入字符i增加的前缀

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<vector>
#include <sstream>
#include<string>
#include<cstring>
#include<list>
using namespace std;
#define MAXN 200002
#define INF 0x3f3f3f3f
#define M 10007
typedef long long LL;
/*
输出字符串中所有前缀在字符串中出现的次数
ababab
*/
char t[MAXN];
int Next[MAXN],cnt[MAXN];
void kmp_pre(char t[])
{
    int m = strlen(t);
    int j,k;
    j = 0;k = Next[0] = -1;
    while(j<m)
    {
        if(k==-1||t[j]==t[k])
            Next[++j] = ++k;
        else
            k = Next[k];
    }
}
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        int l,ans=0;
        scanf("%d",&l);
        scanf("%s",t);
        kmp_pre(t);
        for(int i=1;i<=l;i++)
        {
            cnt[i] = (cnt[Next[i]]+1);
            ans = (ans+cnt[i])%M;
        }
        printf("%d
",ans);
    }
}