Count the stringTime Limit: 2000/1000 MS (Java/Others)Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 5224Accepted Submission(s): 2454Problem 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
14abab
Sample Output
6
Author
foreverlin@HNU
Source
题目链接：?pid=3336题目大意：给一个字符串，求其每个前缀子串在原串中的匹配次数的和题目分析：好题就是数据太水了，1 5 ababa显然是9，但是很多跑出来8的程序都AC了，我来说下正解，，要理解next数组的含义，next数组的值就是当前子串的后缀与前缀匹配的个数，我们先初始化ans＝len，表示每个前缀都只出现了一次，然后根据next数组，找到一个匹配的后缀ans＋1，接着对next回溯一直到为0，相当于把子串里蕴含的前缀子串也找出来，举个例子： s a b a b a i 0 1 2 3 4 5next[i] -1 0 0 1 2 3开始 ans=5next[3] = 1，ans + 1 = 6，next[1] = 0next[4] = 2，ans + 1 = 7，next[2] = 0next[5] = 3，ans + 1 = 8，next[3] = 1，ans + 1 = 9（这里就表示在aba这个后缀里还蕴涵着一个前缀a）所以最后答案是9#include <cstdio>#include <cstring>int const MAX = 1e8;int const MOD = 10007;int next[MAX];char s[MAX];void get_next(){int i = 0, j = -1;next[0] = -1;while(s[i] != '\0'){if(j == – 1 || s[i] == s[j]){i++;j++;next[i] = j;}elsej = next[j];}}int main(){int T, len, ans;scanf("%d", &T);while(T–){memset(next, 0, sizeof(next));scanf("%d %s", &len, s);get_next();ans = len;for(int i = 1; i <= len; i++){int tmp = next[i];while(tmp){ans = (ans + 1) % MOD;tmp = next[tmp];}}printf("%d\n", ans);}}
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