Count the string(hdu3336)

Count the string

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7190    Accepted Submission(s): 3318

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

思路：KMP（next数组）+DP；

dp[i]=(dp[next2[i]]+1)%mod;dp转移方程；

表示以i结尾时所有与[1，i] 不同前缀新增加的个数。aaaaa

dp[1]=1;dp[2]=2(表示新增 a aa)；dp[3]=3（aaa，aa，a).......

dp[i]=(dp[next2[i]]+1)的原因是比如原来是abaab,假设前面的已经算完，现在增加一个字母，

abaaba,next[i]=3;所以新增的和为一个aba符合要求，也就是dp[3];还有一个为新串abaaba;

#include<stdio.h>
#include<algorithm>
#include<iostream>
#include<string.h>
#include<stdlib.h>
#include<queue>
#include<cstdio>
#include<math.h>
void next1(int k);
char str[300000];
int next2 [300000];
char strr[300000];
int dp[300000];
const int mod= 10007;
using namespace std;
int main(void)
{
    int i,j,k,p,q;
    scanf("%d",&k);
    while(k--)
    {
        int sum=0;
        scanf("%d",&p);
        scanf("%s",str);
        int l=strlen(str);
        for(i=0; i<l; i++)
            {strr[i+1]=str[i];}
        next1(l);
        for(i=1; i<=l; i++)
        {
            dp[i]=(dp[next2[i]]+1)%mod;
            sum=(sum+dp[i])%mod;
        }
        printf("%d\n",sum);
    }
    return 0;
}

void next1(int k)
{
    int i,j;
    next2[0]=0;
    next2[1]=0;
    j=0;int cnt=0;
    for(i=2; i<=k; i++)
    {
        while(j>0&&strr[j+1]!=strr[i])
        {
            j=next2[j];
        }
        if(strr[j+1]==strr[i])
        {
            j++;
        }
        next2[i]=j;cnt++;
    }
}