Count the string
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3637 Accepted Submission(s): 1689
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.
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.
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
思路：KMP，但要对其进行变形，当找到失败位置时，要继续考察该位置，一直向前找到字符串首不能再向前找，因为我们不只要计算该子串本身，我们还要计算该子串包含的其他子串，因为这些子串都是原串的子串，这是显然的。个人认为属于KMP的进阶应用，不只是模板题，KMP有待更深入的理解。
AC代码：
1 #include<stdio.h> 2 #include<string.h> 3 int fail[200005]; 4 int sum[200005]; 5 char str[200005]; 6 int T, m, max; 7 void getfail() 8 { 9 fail[0] = -1; 10 int i, j, temp; 11 for(i = 1, j = -1; i < m; i ++) 12 { 13 while(j >= 0 && str[j + 1] != str[i]) 14 { 15 j = fail[j]; 16 } 17 if(str[j + 1] == str[i]) 18 j ++; 19 fail[i] = j; 20 if(j >= 0) 21 { 22 sum[j] ++; 23 if(max < j) 24 max = j; 25 temp = j; 26 while(temp >= 0) 27 { 28 temp = fail[temp]; 29 sum[temp] ++; 30 } 31 } 32 } 33 } 34 int main(int argc, char const *argv[]) 35 { 36 int i, cnt; 37 scanf("%d", &T); 38 while(T--) 39 { 40 memset(str, 0, sizeof(str)); 41 memset(sum, 0, sizeof(sum)); 42 scanf("%d", &m); 43 cnt = m % 10007; 44 max = 0; 45 getchar(); 46 fgets(str, m+1,stdin); 47 getfail(); 48 for(i = 0; i <= max; i ++) 49 { 50 /* printf("OK "); */ 51 sum[i] %= 10007; 52 cnt += sum[i]; 53 cnt %= 10007; 54 } 55 printf("%d ", cnt); 56 } 57 return 0; 58 }