HDU-3336

Count the string

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5512    Accepted Submission(s): 2596

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

 

/**
          题意：给一个字符串，问字符串前缀的个数
          做法：kmp 每次next[]数组记录的下标  前缀数组
**/
#include <iostream>
#include <stdio.h>
#include <cmath>
#include <algorithm>
#include <string.h>
#define maxn 200000 + 10
#define mod 10007
using namespace std;
char ch[maxn];
int mmap[maxn];
int Next[maxn];
void getnext(char *p)
{
          int len = strlen(p);
          int j,k;
          j=0;
          k = -1;
          Next[0] = -1;
          while(j < len)
          {
                    if(k == -1 || p[j] == p[k])
                    {
                              j++;
                              k++;
                              Next[j] = k;
                    }
                    else k = Next[k];
          }
}
int main()
{
#ifndef ONLINE_JUDGE
          freopen("in.txt","r",stdin);
#endif // ONLINE_JUDGE
          int T;
          scanf("%d",&T);
          while(T--)
          {
                    int n;
                    scanf("%d",&n);
                    scanf("%s",ch);
                    getnext(ch);
                    mmap[0] = 0;
                    int res= 0 ;
                    for(int i=1;i<=n;i++)
                    {
                              mmap[i] = mmap[Next[i]] + 1;
                              mmap[i] %= mod;
                              res +=  mmap[i];
                              res %= mod;
                    }
                    printf("%d\n",res);
          }
          return 0;
}