hdu 3336 Count the string【kmp】

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本文介绍了一种使用KMP算法解决字符串问题的方法，具体为计算一个字符串的所有非空前缀在其自身中出现的总次数，并通过一个示例程序详细展示了算法的具体实现。

Count the string

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 11717 Accepted Submission(s): 5461

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

Author

foreverlin@HNU

题意：给你一个字符串，问字符串每一个前缀在字符串中的出现总次数。

思路：kmp的应用，自身和自身进行匹配，每次匹配时，如果没有匹配到结束，模式串按next数组向后移动，出现匹配至结束的情况，匹配串往后移动一位

　　，匹配过程中，出现失配时，统计当前已经匹配的字符个数，累加到总数中，直到匹配串移动到最后一位。

其实前两天就在网上看题解用kmp+dp的方法来写了这道题，然而后来细思一些细节，自己并没有弄懂，所以，我就没有写成题解，最后，还是我师父出面，把这道题给解释了

#include<stdio.h>
#define N 200010
#define mod 10007
int next[N];
char str[N];
void Get_next(int n)
{
    int j ,k;
    j = 0;
    next[j] = k = -1;
    while(j <= n)
    {
        if(k == -1||str[j] ==str[k])
            next[++j] = ++k;
        else
            k = next[k];
    }
    return;
}

int KMP(int n)
{
    int i,j,k;
    i = 1;
    j = k = 0;
    while(i <= n)
    {
        if(j == -1||str[i]==str[j])
        {
            i++;
            j++;
        }
        else//失配时 
        {
            k += j;//每次累加起点不同匹配的字符串长度，保证了起点不重复 
            k%=mod;
            j = next[j];
        }
    }
    return k ;
}
int main()
{
    int t,n;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        getchar();
        scanf("%s",str);
        Get_next(n);
        printf("%d\n",(KMP(n)+n)%mod);
    }
    return 0;
}