hdu 3336 Count the string

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=3336

分析：求字符串中所有前缀的匹配次数，先求出next数组，从n->1开始枚举计数。

/*Count the string

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5374    Accepted Submission(s): 2538


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
 
*/
#include <cstdio>
#include <cstring>
#define MOD 10007;
const int maxn = 200000 + 10;
int n, next[maxn];
char s[maxn];

void getNext()
{
    next[0] = -1;//初始化 
    int j = 0, k = -1;
    while(j < n){//遍历字符串 
        if(k == -1 || s[j] == s[k]) next[++j] = ++k;
        else k = next[k];
    }
}

int main()
{
    int t;
    scanf("%d", &t);
    while(t--){
        scanf("%d", &n);
        scanf("%s", s);
        getNext();
        int cnt = 0;
        for(int i = n; i > 0; i--){//从n开始匹配 
            //每一步都要取模，以防止溢出 
            cnt = (cnt+1) % MOD;
            int j = next[i];
            while(j){
                cnt = (cnt+1)%MOD;
                j = next[j]; //沿着next向下匹配 
            }
        }
        printf("%d\n", cnt);
    }
    return 0;
}