Count the string

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

14abab

Sample Output

6

题意：一个字符串的子串在该字符串出现的次数之和

       例如：

              字符串abab

            子串： a , ab , aba , abab

  出现的次数：2 + 2  +  1   +  1    =  6

如果用循环来做的话，N比较大，所以会超时，那应该怎么做呢 ？

这里就用到了next[ ]的特点了，利用DP , 看来DP学不好也不行啊。。。。

 next[ ]表示的是最大前缀和 和 后缀和的字符串长度

         dp[ i ] = dp[ next[i] ] + 1;

 也就是说next[ i ] 不等于0的话

         那么子串出现的次数至少有2次

#include <iostream>#include <cstdio>#include <cstring>#define Mod 10007const int MAX=2e5+10;char str[MAX] ;int next[MAX] , dp[MAX];void getnext(int n){  int j = -1 , i =0;  next[0] = -1;   while(i < n)   {       if( j == -1 || str[i] == str[j])         {                 i++;j++;next[i] = j;   }else{ j = next[j];}   }}int main(){       int t;       scanf("%d",&t);    while(t--)   {        int n;        scanf("%d",&n);          scanf("%s",str);             getnext(n);             memset(dp , 0 , sizeof(dp));             int sum = 0;             for(int i = 1; i <= n; i++)               {                  dp[i] = dp[next[i]] + 1;                  sum += dp[i]%Mod;   }   printf("%d\n",sum%Mod);   }    return 0;}