关于next数组的一个例题

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本文探讨了一种使用next数组来高效计算字符串所有前缀出现次数之和的方法，通过实例解析了算法流程，并提供了完整的C++代码实现。

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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

就是问你前缀出现的次数之和

这里利用next数组，例如ababca这个数组，next数组从0到n值分别为-1 0 0 1 2 0 1，这里的next数组的意思可以为当前字符/字符串在从开头到哪个位置结束之间的距离，也可以表示为出现次数，再把本身的匹配加上（也就是字符串长度）就是答案了。

#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=0; i<=n; i++)
        {
            cout<<Next[i]<<" dddd";
        }
        cout<<endl;
        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;
}