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：求一个串中所有的前缀各自在串中出现次数的和。

分析：先来看看一个例子

i      1  2  3  4  5  6

str[i]   a  b  a  b  a  b

next[i] 0  0  1  2  3  4

next[i]=j; 表示从1----j 和 i-j+1----i这一段是相同的，利用next数组，依次推导

例如： i=5时 ababa  所包含的 前缀数 等于 以第3个a结尾的前缀数(即ababa本身) + （next[5]=3）str[3]所包含的前缀数量（即aba包含的前缀数量）

于是可以得到递推方程 dp[j]=dp[next[j]]+1  dp[0]=0;

对于这个例子，可以有：

dp[1]=dp[n[1]]+1=1;  a

dp[2]=dp[n[2]]+1=1;  ab

dp[3]=dp[n[3]]+1=2; a aba

dp[4]=dp[n[4]]+1=2; ab abab

dp[5]=dp[n[5]]+1=3; a aba ababa

dp[6]=dp[n[6]]+1=3; ab abab ababab

#include <iostream>
#include <stdio.h>
#include <string>
#include <cstring>
#include <cmath>
#include <algorithm>

using namespace std;
char s[222222];
int n;
int nt[222222];
int dp[222222];

void getnext()
{
    int k=-1,j=0;
    nt[0]=-1;

    while(j<n)
    {
        if(k<0 || s[j]==s[k])
        {
            j++;
            k++;
            nt[j]=k;
        }
        else
            k=nt[k];
    }
}

int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        scanf("%s",s);

        getnext();

//        for(int i=0;i<=n;i++)
//            cout<<nt[i]<<" ";
//        cout<<endl;
     for(int i=1;i<=n;i++) dp[i]=1;
     dp[0]=0;

     int ans=0;
     for(int i=1;i<=n;i++)
     {
         dp[i]=dp[nt[i]]+1;
         ans=(ans+dp[i])%10007;
     }
     printf("%d\n",ans);

    }
    return 0;
}

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