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
14abab
Sample Output
6
这道题可以用kmp做，，next[i]表示当前的字符长度下前缀与后缀相同的最大长度，其实就是和前面的next[i]个字符重复，所以要加上前面出现过的总次数sum[next[i]]。
#include<stdio.h>#include<string.h>int sum[200006],n,next[200006];char s[200006];void nextt(){int i,j;i=0;j=-1;memset(next,-1,sizeof(next));while(i<n){if(j==-1 || s[i]==s[j]){i++;j++;next[i]=j;}else j=next[j];}}int main(){int m,i,j,T,num;scanf("%d",&T);while(T–){scanf("%d%s",&n,s);nextt();num=0;for(i=1;i<=n;i++){sum[i]=1;sum[i]=(sum[i]+sum[next[i]])%10007;num=(num+sum[i])%10007;}printf("%d\n",num);}return 0;}
有山就有路，有河就能渡。
