HDU：3336-Count the string（next数组理解）

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

解题心得：

题意就是给你一个字符串，问你字符串中和每个前缀匹配的子串有多少个，可以前缀和前缀自身匹配。

其实就是一个对于next数组的理解，next数组记录的就是当前字符为后缀匹配的前缀，所以很简单，直接next匹配的时候记录总数就可以了。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<queue>
#include<vector>
#include<stack>
#include<string>
using namespace std;
const int maxn = 2e5+100;
const int MOD = 10007;
char s[maxn];
int Next[maxn];

int cal_next(int n){
int ans = 1;
int k = -1;
Next[0] = -1;
for(int i=1;i<n;i++){
ans++;//自身和自身匹配
ans %= MOD;
while(k > -1 && s[k+1] != s[i])
k = Next[k];
if(s[k+1] == s[i]){
k++;
ans++;
ans %= MOD;
}
Next[i] = k;
}
return ans%MOD;
}

int main() {
int t;
scanf("%d",&t);
while(t--){
int n;
scanf("%d",&n);
scanf("%s",s);
int ans = cal_next(n);
printf("%d\n",ans);
}
return 0;
}