HDU3336--所以前缀的出现次数和（KMP）

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

rt

思路：https://www.cnblogs.com/Tree-dream/p/7443897.html

差不多了，dp只是简化递归而已。

#define IOS ios_base::sync_with_stdio(0); cin.tie(0);
#include <cstdio>//sprintf islower isupper
#include <cstdlib>//malloc  exit strcat itoa system("cls")
#include <iostream>//pair
#include <fstream>//freopen("C:\Users\13606\Desktop\Input.txt","r",stdin);
#include <bitset>
//#include <map>
//#include<unordered_map>
#include <vector>
#include <stack>
#include <set>
#include <string.h>//strstr substr strcat
#include <string>
#include <time.h>// srand(((unsigned)time(NULL))); Seed n=rand()%10 - 0~9;
#include <cmath>
#include <deque>
#include <queue>//priority_queue<int, vector<int>, greater<int> > q;//less
#include <vector>//emplace_back
//#include <math.h>
#include <cassert>
#include <iomanip>
//#include <windows.h>//reverse(a,a+len);// ~ ! ~ ! floor
#include <algorithm>//sort + unique : sz=unique(b+1,b+n+1)-(b+1);+nth_element(first, nth, last, compare)
using namespace std;//next_permutation(a+1,a+1+n);//prev_permutation
//******************
clock_t __START,__END;
double __TOTALTIME;
void _MS(){__START=clock();}
void _ME(){__END=clock();__TOTALTIME=(double)(__END-__START)/CLOCKS_PER_SEC;cout<<"Time: "<<__TOTALTIME<<" s"<<endl;}
//***********************
#define rint register int
#define fo(a,b,c) for(rint a=b;a<=c;++a)
#define fr(a,b,c) for(rint a=b;a>=c;--a)
#define mem(a,b) memset(a,b,sizeof(a))
#define pr printf
#define sc scanf
#define ls rt<<1
#define rs rt<<1|1
typedef pair<int,int> PII;
typedef vector<int> VI;
typedef unsigned long long ull;
typedef long long ll;
typedef double db;
const db E=2.718281828;
const db PI=acos(-1.0);
const ll INF=(1LL<<60);
const int inf=(1<<30);
const db ESP=1e-9;
const int mod=(int)1e9+7;
const int N=(int)1e6+10;

int Next[N],lenb;
char b[N];

void set_naxt()                  //子串的next数组:一个字符串的最长前缀和最长后缀相同的长度
{
    int i=0,j=-1;
    Next[0]=-1;                  //初始化为-1，表示不存在相同的最长前缀和最大后缀
    while(i<lenb)
    {
        if(j==-1||b[i]==b[j])
        {
            i++;
            j++;
            Next[i]=j;           //把j（相同的最大前缀和最大后缀的长）赋值next[i]
        }
        else
            j=Next[j];           //往前回溯
    }
}
int dp[N];

void solve()
{
    sc("%d",&lenb);
    sc("%s",b);
    b[lenb++]='%';
    fo(i,0,lenb)dp[i]=0;
    set_naxt();
    int ans=0;
    for(int i=1;i<lenb;++i)
        dp[i]=dp[Next[i]]+1,ans+=dp[i],ans%=10007;
    pr("%d
",ans);
}

int main()
{
    int T;
    sc("%d",&T);
    while(T--)solve();
    return 0;
}

/**************************************************************************************/