Count the string

最新推荐文章于 2020-11-18 21:41:36 发布

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最新推荐文章于 2020-11-18 21:41:36 发布

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分类专栏： ACM 文章标签： KMP

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本文介绍了一种基于KMP算法解决字符串匹配问题的方法，通过计算字符串所有前缀的出现次数之和，并给出具体实现代码。

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<span style="font-size: 18px; text-align: justify; font-family: Arial, Helvetica, sans-serif; background-color: rgb(255, 255, 255);">Count the string</span>

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

abab

Sample Output

题意：给定一字符串，求它所有的前缀出现的次数的和。

在kmp算法中，利用到的next数组很神奇。next数组中求得是当前最大的前缀等于后缀的前缀子串的位置（也是子串最长公共前后缀串的长度）。

如果用dp[i]表示该字符串前i个字符中出现任意以第i个字符结尾的前缀的次数，它的递推式是 dp[i]=dp[next[i]]+1,即以第i个字符结尾的前缀数等于以第next[i]个字符为结尾的前缀数加上它自己本身，这里要好好理解一下，不太好解释。

举个例子：

i 1 2 3 4 5 6

字符串 a b a b a b

dp[i] 1 1 2 2 3 3

aba中出现的前缀为a,aba,所以dp[3]是2，ababa中出现的前缀为a,aba,ababa,所以dp[5]是3，当i=5时,next[5]=3，所以dp[i]=dp[next[i]]+1

#include<string.h>
#include<stdio.h>
char B[200010];//A是主串
int next[200010];
int dp[200010];
void snext(char P[],int next[])
{
    int i;
    int m=strlen(P);
    next[0]=0;
    int k=0;
    for(i=1;i<m;i++)
    {
        while(k>0&&P[i]!=P[k])
            k=next[k-1];
        if(P[i]==P[k])
            k++;
        next[i]=k;
    }
}
int main()
{
     int i,T,l,sum;
     scanf("%d",&T);
     while(T--)
     {
        sum=0;
      memset(next,0,sizeof(next));
       memset(dp,0,sizeof(dp));
      scanf("%d",&l);
      scanf("%s",B);
      snext(B,next);
      for(i=0;i<l;i++)
        {
           dp[i]=(dp[next[i]-1]+1)%10007;
           sum=(sum%10007+dp[i]%10007)%10007;
        }
        printf("%d\n",sum);
     }
      return 0;
}