HDU3336 Count the string（KMP+DP）

计算字符串的每个前缀在字符串里出现的次数。

一个KMP+DP，在KMP的next数组中，next[i]=j就表示在前i个字符串当中，从前1到j的j长度字符字串和i-j+1到i的j长度的字符字串相等。那么，每一个位置的字符只要继承它的next[i]位置的数量，再加一就是它当前的数量了。

#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;
const int N=200005;
const int MOD=10007;

char str
;
int nextt
;
int dp
;

void makenextt(char p[],int len)
{
nextt[0]=0;
for(int i=1,k=0;i<len;i++)
{
nextt[0]=0;
while(k>0&&p[k]!=p[i])
k=nextt[k-1];
if(p[k]==p[i])
k++;
nextt[i]=k;
}
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
int n;
scanf("%d",&n);
scanf("%s",str);
makenextt(str,n);
int ans=0;
for(int i=0;i<n;i++)
{
if(nextt[i]==0)
dp[i]=1;
else
dp[i]=(dp[nextt[i]-1]+1)%MOD;
ans+=dp[i];
ans%=MOD;
}
printf("%d\n",ans);
}
return 0;
}

Count the string

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

Total Submission(s): 12612    Accepted Submission(s): 5817


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

14abab

 

Sample Output

6