大家好，又见面了，我是你们的朋友全栈君。
http://acm.hdu.edu.cn/showproblem.php?pid=3336
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
next数组的值就是记录当前位置向前多少个和这个字符串开头多少个相等，有了这样我们只要记录每个next这对应有多少个，然后加上n个自身的一个匹配就行了。
这个题目的关键还是对kmp的理解，本质上kmp的next就是表示串内关系的一个值。
#include <iostream>
#include <string.h>
#include <stdio.h>
using namespace std;
char str[200001];
int next[200001];
int rec[200001];
int n;
void get_next()
{
int i=0,j=-1;
next[0]=-1;
while(str[i])
{
if(j==-1 || str[i]==str[j])
next[++i]=++j;
else
j=next[j];
}
}
int main()
{
int t,i,j;
int ans;
scanf("%d",&t);
while(t--)
{
ans=0;
scanf("%d",&n);
scanf("%s",str);
memset(rec,0,sizeof(rec));
str[n]='a';//next[0]是-1，可看做补0的空缺。
get_next();
for(i=1; i<=n; i++)
rec[next[i]]++;
for(i=1; i<=n; i++)
if(rec[i] > 0)
{
ans+=rec[i];
ans%=10007;
}
ans+=n;
printf("%d\n",ans%10007);
}
return 0;
}
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