Count the string

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

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

6

题意：

比如abab中a，ab，aba，abab在原串中的个数的总和为2+2+1+1 = 6。（举个例子应该更好理解）

思路：

首先就是求出Next数组，这样才知道可以用动态规划，设dp[i]表示子串mo[0~i]共含有以mo[i]为结尾的前缀的数目，则以mo[i]结尾的前缀数就是自己本身加上以mo[Next[i]]结尾的前缀数,然后还要取模。

#include <iostream>
#include <cstdio>
#include <cstring>
const int maxn = 2e+5 + 10;
const int mod = 10007;
char mo[maxn];
int Next[maxn], dp[maxn];
void GetNext() {
    int i = 0, j = -1, len = strlen(mo);
    while (i < len) {
        if (j == -1 || mo[i] == mo[j]) Next[++i] = ++j;
        else j = Next[j];
    }
}
int main() {
    int t, n;
    scanf("%d", &t);
    while (t--) {
        memset(dp, 0, sizeof(dp));
        scanf("%d", &n);
        scanf("%s", mo);
        Next[0] = -1;
        GetNext();
        int sum = 0;
        for (int i = 1; i <= n; i++) {
            dp[i] = dp[Next[i]] + 1;
            sum = (dp[i] + sum) % mod;
        }
        printf("%d\n", sum);
    }
    return 0;
}