hdu 3336 Count the string

Count the string

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

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

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

題目大意：求字元串字首在字元串中出現的次數和

KMP+DP

首先求出字元串的next數組

狀态轉移方程：dp[i]+=dp[next[i]]+1,ans+=dp[i]

如何了解？

例：    a b a b a

next：0 0 1 2 3

dp：   1 1 2 2 3

對于+1，因為本身和本身比對

 dp[3]=dp[1]+1 

因為next[3]=1,是以第1個和第3個是相同的

第1個也會出現在第3個中

隻不過這裡把它加到了第3個裡

也就是說：

dp[i]=dp[j]+1

[1,j]=[i-j+1,i]

是以[1,j]中出現過的在[i-j+1,i]中重複出現

次數本應累加到[i,j]裡，實際累加到了[i-j+1,i]裡 

　　

作者：xxy

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