手动转田神的大作：

 

D. Prefixes and Suffixes

time limit per test

1:second

memory limit per test:

256 megabytes

input:

standard input

output:

standard output

You have a string s = s1s2...s|s|, where |s| is the length of string s, and si its i-th character.

Let's introduce several definitions:

• A substring s[i..j] (1 ≤ i ≤ j ≤ |s|) of string s is string sisi + 1...sj.
• The prefix of string s of length l (1 ≤ l ≤ |s|) is string s[1..l].
• The suffix of string s of length l (1 ≤ l ≤ |s|) is string s[|s| - l + 1..|s|].

Your task is, for any prefix of string s which matches a suffix of string s, print the number of times it occurs in string s as a substring.

Input

The single line contains a sequence of characters s1s2...s|s| (1 ≤ |s| ≤ 105) —  string s. The string only consists of uppercase English letters.

Output 

In the first line, print integer k (0 ≤ k ≤ |s|) — the number of prefixes that match a suffix of string s. Next print k lines, in each line print two integers lici. Numbers li ci mean that the prefix of the length li matches the suffix of length li and occurs in string s as a substring ci times. Print pairs li ci in the order of increasing li.

Sample test(s)

input

ABACABA

output

31 43 27 1

input

AAA

output

31 32 23 1

 

题目链接 :

 

题目大意 :给一个字符串，求其前缀等于后缀的子串，输出子串的长度和其在原串中出现的次数，子串长度要求曾序输出

 

题目分析 :首先得到母串的next数组，next数组的含义是next[j]的值表示str[0...j-1]（我的next[0]是-1）这个子串的前后缀匹配的最长长度，如样例1

index  0  1  2  3  4  5  6  7

str    A  B  A  C  A  B  A

next   -1 0  0  1  0  1  2  3

next[6] = 2即ABACAB这个子串的前后缀最长匹配是2(AB)

由此性质我们可以发现满足条件的子串即是next[next[len。。。]]不断往前递归直到为0，因为长的可能会包含短的，我们可以递归得到re数组(re记录的就是子串出现的次数)re数组的递归式为re[ next[temp] ] += re[temp];

 

 

1 #include 
         
           2 #include 
          
            3 int const MAX = 1e5 + 2; 4 char str[MAX]; 5 int next[MAX], re[MAX], le[MAX], sub[MAX];; 6 int len, count; 7  8 void get_next() 9 {10     int i = 0, j = -1;11     next[0] = -1;12     while(str[i] != '\0')13     {14         if(j == -1 || str[i] == str[j])15         {16             j++;17             i++;18             next[i] = j;19         }20         else21             j = next[j];22     }23 }24 25 void kmp()26 {27     get_next();28     len = strlen(str);29     memset(re,0,sizeof(re));30     for(int i = 1; i <= len; i++)31         re[i] = 1;32     int temp = len;33     for(; temp > 0; temp--)34         if(next[temp])35             re[ next[temp] ] += re[temp];36     count = 0;37     le[count] = len;38     sub[count++] = 1;39     len = next[len];40     while(len)41     {42         le[count] = len;43         sub[count++] = re[len];44         len = next[len];45     }46 }47 48 int main()49 {50     while(scanf("%s", str) != EOF)51     {52         kmp();53         printf("%d\n",count);54         for(int i = count - 1; i >= 0; i--)55             printf("%d %d\n",le[i], sub[i]);56     }57 }