hdu3336 Count the string

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. 

 

KMP理解题目 1 //字符串上KMP(水)

 2 //从前向后扫，失配函数的位置就是一个前缀的位置减1
 3 //加起来就好了
 4 #include<cstring>
 5 #include<cstdio>
 6 #include<algorithm>
 7 using namespace std;
 8 const int MAX = 200000;
 9 const int MOD = 10007;
10 char str[MAX];
11 int next[MAX],vis[MAX];
12 int main()
13 {
14     int cas,n;
15     scanf("%d",&cas);
16     while(cas--)
17     {
18         scanf("%d %s",&n,str);
19         next[0]=next[1]=0;
20         for(int i=1;i<n;i++)
21         {
22             int j=next[i];
23             while(j&&str[i]!=str[j]) j=next[j];
24             if(str[i]==str[j])
25             next[i+1]=j+1;
26             else next[i+1]=0;
27         }
28         int ans=0,cnt=0;
29         for(int i=0;i<n;i++)
30         {
31             if(next[i])
32             {
33             //    cnt++;
34                 ans=(ans+2)%MOD;
35             }
36             else
37             ans=(ans+1)%MOD;
38         }
39         if(next[n]) ans=(ans+1)%MOD;
40         printf("%d\n",(ans)%MOD);
41     }
42     return 0;

43 }