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.

题意：求一个字符串和它所有前缀的匹配次数总和，如'abab'与前缀'ab'匹配次数为2。

用扩展KMP与自身匹配。

 1 #include<stdio.h>
 2 #include<string.h>
 3 #include<algorithm>
 4 using namespace std;
 5 
 6 const int maxn=2e5+5;
 7 const int mod=1e4+7;
 8 
 9 char s[maxn];
10 int ext[maxn],nxt[maxn];
11 
12 void EKMP(char s[],char t[],int lens,int lent){
13     int i,j,p,l,k;
14     nxt[0]=lent;j=0;
15     while(j+1<lent&&t[j]==t[j+1])j++;
16     nxt[1]=j;
17     k=1;
18     for(i=2;i<lent;i++){
19         p=nxt[k]+k-1;
20         l=nxt[i-k];
21         if(i+l<p+1)nxt[i]=l;
22         else{
23             j=max(0,p-i+1);
24             while(i+j<lent&&t[i+j]==t[j])j++;
25             nxt[i]=j;
26             k=i;
27         }
28     }
29 
30     j=0;
31     while(j<lens&&j<lent&&s[j]==t[j])j++;
32     ext[0]=j;k=0;
33     for(i=1;i<lens;i++){
34         p=ext[k]+k-1;
35         l=nxt[i-k];
36         if(l+i<p+1)ext[i]=l;
37         else{
38             j=max(0,p-i+1);
39             while(i+j<lens&&j<lent&&s[i+j]==t[j])j++;
40             ext[i]=j;
41             k=i;
42         }
43     }
44 }
45 
46 int main(){
47     int T;
48     scanf("%d",&T);
49     while(T--){
50         int n;
51         scanf("%d%s",&n,s);
52         EKMP(s,s,n,n);
53         int cnt=0;
54         for(int i=0;i<n;++i){
55             cnt=(cnt+ext[i])%mod;
56         }
57         printf("%d\n",cnt);
58     }
59     return 0;
60 }
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