hdu 3336 Count the string kmp

Count the string

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

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

  

题意：求出前n个字符组成的字符串能被原串匹配次数之和

如 ababababab

a=5

ab=5

aba=4

abab=4

ababa=3

ababab=3

abababa=2

abababab=2

ababababa=1

ababababab=1

sum=5+5+4+4+3+3+2+2+1+1=30

对于任意一个k>=1，每往前找到next[k]的位置，k与next[k]之间的字符串就是当前考虑的在找匹配的字符串，也就是说，每到一个k 能往前面找到多少次next[k]那么该字符串就是被匹配了多少次

#include<cstdio>
#include<iostream>
using namespace std;
char a[200010];
int nxt[200010];  
int main(){
	int n;
	scanf("%d",&n);
	while(n--){
		int m,sum=0;
		scanf("%d%s",&m,a);
		int i=-1,j=0;  
		nxt[0]=-1;  
    	        while(j<m){  
        	        if(i==-1||a[i]==a[j])nxt[++j]=++i;  
        	        else i=nxt[i];  
 		 } 
		for(i=1;i<=m;i++){
			j=i;
			sum++;
			while(nxt[j]!=0){
				sum=(sum+1)%10007;
				j=nxt[j];
			}	
		}
		printf("%d\n",sum);
	}
  return 0；	
}