HDU3336Count the string

HDU3336Count the string

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

6

 

Author

foreverlin@HNU

 

Source

HDOJ
Monthly Contest – 2010.03.06

 

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#include<cstdio>
#include<cstring>
#define mod 10007

using namespace std;

int T,l,next[],ans,num[];
char s[];

int main() {
    scanf("%d",&T);
    while(T--) {
        memset(num,,sizeof(num));
        memset(next,,sizeof(next));
        scanf("%d",&l);
        scanf("%s",s);
        ans=;
        for(int i=,j=; i<l; i++) {
            while(j&&s[i]!=s[j]) j=next[j];
            if(s[i]==s[j]) ++j;
            next[i+]=j;
        }
        for(int i=; i<=l; i++) {
            num[i]+=(num[next[i]]+)%mod;
            ans=(ans+num[i])%mod;
        }
        printf("%d\n",ans);
    }
    return ;
}