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

Count the string

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6672    Accepted Submission(s): 3089

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

dp[i] = dp[Next[i]] + 1;

但是dp[Next[i]] = dp[dp[Next[i]]+1]+1;

感觉有点像递归的思想，dp[i] 加 1 是算它本身，然而dp[Next[i]]代表的是长度为 i 的最大匹配（前缀和后缀的最大匹配）在母串中的个数，然后层层找出所有的个数

例：

为什么要用Next呢，因为要看后缀

dp[1] 代表的是长度为1的串中分别有1个 a

dp[2] 代表的是长度为2的串中分别有1个 ab

dp[3] 代表的是长度为3的串中分别有1个 aba, a

dp[4] 代表的是长度为4的串中分别有1个 abab, ab

dp[5] 代表的是长度为5的串中分别有1个 ababa, aba, a

#include<iostream>
#include<stdio.h>
#include<string.h>
using namespace std;

#define MOD 10007
#define N 210000

char M[N];
int Next[N], dp[N];

void FindNext(int len)
{
    int i=0, j=-1;
    Next[0] = -1;

    while(i<len)
    {
        if(j==-1 || M[i]==M[j])
            Next[++i] = ++j;
        else
            j = Next[j];
    }
}

int main()
{
    int t;
    scanf("%d", &t);
    while(t--)
    {
        int n, i, Sum=0;
        memset(dp, 0, sizeof(dp));

        scanf("%d%s", &n, M);

        FindNext(n);
        for(i=1; i<=n; i++)
        {
            dp[i] = (dp[Next[i]] + 1)%MOD;
            Sum = (dp[i]+Sum)%MOD;
        }

        printf("%d\n", Sum);
    }
    return 0;
}