K - Count the string HDU - 3336

题目链接：https://vjudge.net/contest/70325#problem/K

题目：

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
题意：给你一个字符串，求出所有前缀的个数和，可以重叠
思路：首先前缀个数就是字符串长度len，然后利用kmp求next数组，如果next[n]为4，对应前后缀最长相等的部分为abcd，那么a,ab,abc,abcd,都要加一次就是加next【n】,所以这题就是相当于
求next数组，只要满足next[i]+1!=next[i+1]就行，就可以找到所有前缀的个数和了

// 
// Created by HJYL on 2019/8/16.
//
#include <iostream>
#include <vector>
#include <map>
#include <string>
#include <queue>
#include <stack>
#include <set>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
using namespace std;
typedef long long ll;
const int maxn=1e6+10;
char str[maxn];
int nextt[maxn];
void getnext()
{
    int i=0,j=-1;
    nextt[0]=-1;
    int len=strlen(str);
    while(i<len)
    {
        if(j==-1||str[i]==str[j])
        {
            i++,j++;
           //if(str[i]!=str[j])
                nextt[i]=j;
           // else
               // nextt[i]=nextt[j];
        } else
            j=nextt[j];
    }
}
int main()
{
    //freopen("C:\\Users\\asus567767\\CLionProjects\\untitled\\text","r",stdin);
    int T;
    scanf("%d",&T);
    int n;
    while(T--)
    {
        scanf("%d",&n);
        getchar();
        scanf("%s",str);
        int len=strlen(str);
        //cout<<"len="<<len<<endl;
        getnext();
        int res=len+nextt[len];
        for(int i=0;i<len;i++)
        {
            if(nextt[i]+1!=nextt[i+1]&&nextt[i]>0)
                res+=nextt[i];
        }
        printf("%d\n",res%10007);
    }
    return 0;
}