hdu 3336 Count the string(kmp应用)

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

 

给你一个字符串算这里面所有前缀出现的次数和。比如字符串abab，a出现2次，ab出现2次，aba出现1次，abab出现1次。

1. 
   

   #include <iostream>

   #include <string.h>

   #include <stdio.h>

   using namespace std;

   int Next[200010],dp[200010]; //dp[i]=以第i个字符结尾的串中所有前缀的计数和

   char str1[200010];

   void GetNext(int len,char *s)

   {

       Next[1]=0;

       int i=1;

       int j=0;

       dp[0]=0;

       while (i<=len)

       {

           if(j==0 || s[i]==s[j])

           {

               dp[i]=(dp[j]+1)%10007;

               i++;

               j++;

               if(s[i]==s[j])

                   Next[i]=Next[j];

               else

                   Next[i]=j;

           }

           else

               j=Next[j];

       }

   }

   
   

   int main()

   {

       int t,n,ans;

       scanf("%d",&t);

       while (t--)

       {

           ans=0;

           scanf("%d",&n);

           getchar();

           scanf("%s",str1+1);

           GetNext(n,str1);

           for (int i=1; i<=n; i++)

           {

               ans=(ans+dp[i])%10007;

           }

           

           printf("%d\n",ans);

       }

       return 0;

   }