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HDU 3336 Count the string（kmp）

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HDU 3336 Count the string（kmp）

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
Source HDOJ Monthly Contest – 2010.03.06

我看完別人博客後發現了kmp的一個秘密，就是當nexti[i]!=0（i從1開始），那麼就會有一個串就是母串從頭開始的一部分，至於為什麼自己把nexti]數組輸出一下吧

代碼還是kmp模板：

#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#define eps 1e-6
#define INF (1<<30)
#define PI acos(-1.0)
using namespace std;

#define N 220000
#define mod 10007

int n,next[N];
char a[N];

void getfail(char *a)
{
   int i,j;
   int len=strlen(a);
   j=-1;
   i=0;
   next[0]=-1;
   while(i