kmp（前缀出现次数next应用）

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

Count the string

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 17068    Accepted Submission(s): 7721

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

 

 

Author

foreverlin@HNU

 

 

Source

HDOJ Monthly Contest – 2010.03.06

 

 

Recommend

lcy

题意：求字符串所有前缀出现的次数（包括本身）

思路：求所有next的值的个数加上前缀本身

 

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <iostream>
#include <algorithm>
#include <iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include <stdio.h>
#include <string.h>
using namespace std;
char a[200009];

int num;


void getnext(char* a, int len , int *next)
{
    next[0] = -1 ;
    int k = -1 , j = 0 ;
    while(j < len)
    {
        if(k == -1 || a[j] == a[k])
        {
            k++;
            j++;
        //    if(a[j] != a[k])
                next[j] = k ;
          //  else
          //  {
          //      next[j] = next[k];
           // }

        }
        else
        {
            k = next[k];
        }
    }
}

int main()
{int n ;
    scanf("%d" , &n);
    while(n--)
    {
        int next[200009];
        int l ;
        scanf("%d" , &l);
        scanf("%s" , a);
        getnext(a , l , next);
        int j = 0 ;
        for(int i = 0 ; i <= l ; i++)
        {
        //    cout << next[i] << " " ;
            j = i ;
            while(next[j] > 0)
            {
                num = (num + 1) % 10007 ;
                j = next[j];
            }
        }
       // cout << endl ;

        printf("%d\n" , (num + l)%10007);
        num = 0 ;
    }

    return 0 ;
}