hdu 3336 Count the string(kmp)

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

题意：给你一个字符串 问你 他的所有前缀 在当前字符串出现的总次数

思路：本来是想 每个前缀都去匹配一次 当然会TLE 所以仔细想想 next数组 不就是前缀数组的匹配情况吗

所以 只要next数组>0我们就可以在总数上+1再加上非真前缀数即可

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<string>
#include<vector>
#include<stack>
#include<bitset>
#include<cstdlib>
#include<cmath>
#include<set>
#include<list>
#include<deque>
#include<map>
#include<queue>
#define ll long long int
using namespace std;
inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
inline ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
int moth[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
int dir[4][2]={1,0 ,0,1 ,-1,0 ,0,-1};
int dirs[8][2]={1,0 ,0,1 ,-1,0 ,0,-1, -1,-1 ,-1,1 ,1,-1 ,1,1};
const int inf=0x3f3f3f3f;
const ll mod=10007;
int nextt[200007];
int n;
void getnext(string s){
    nextt[1]=0;
    for(int i=2,j=0;i<=n;i++){
        while(j>0&&s[i-1]!=s[j]) j=nextt[j];
        if(s[i-1]==s[j]) j++;
        nextt[i]=j;
    }
}
int main(){
    ios::sync_with_stdio(false);
    int t;
    cin>>t;
    while(t--){
        string s;
        cin>>n>>s;
        getnext(s);
        int ans=n; //非真前缀数 
        for(int i=1;i<=n;i++){
            ans=(ans+(nextt[i]==0?0:1))%mod; //如果没有前后缀相同就不处理 
        }
        cout<<ans<<endl;
    }
    return 0;
}