HDU3336-Count the string

Count the string

                                                                            Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

                                                                                                       Total Submission(s): 8965    Accepted Submission(s): 4174

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 [email protected]   Source HDOJ Monthly Contest – 2010.03.06   Recommend lcy

       题意：给以字符串计算出以前i个字符为前缀的字符中在主串中出现的次数和

   解题思路：next[i]记录的是长度为i不包括自身的最大首尾重复子串长度，sum[i]记录长度为next[i]的前缀所重复出现的次数  

#include <iostream>
#include <stdio.h>
#include <queue>
#include <cstring>
#include <stack>

using namespace std;

char a[200009];
int nt[200009];
int sum[200009];
int n;

void get_next()
{
    nt[0]=-1;
    for(int i=0; i<n; i++)
    {
        int k=nt[i];
        while(k>=0&&a[i]!=a[k])
            k=nt[k];
        nt[i+1]=k+1;
    }
}

int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        scanf("%s",a);
        get_next();
        int ans=0;
        sum[0]=0;
        for(int i=1;i<=n;i++)
        {
            sum[i]=(sum[nt[i]]+1)%10007;
            ans+=sum[i];
            ans%=10007;
        }
        printf("%d\n",ans);
    }
    return 0;
}