hdu3336—Count the string（kmp+dp）

Count the string

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 11169    Accepted Submission(s): 5207

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
    
    
    
    

 

Author

foreverlin@HNU

 

Source

HDOJ Monthly Contest – 2010.03.06

 

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lcy

 

Statistic |  Submit |  Discuss | Note

题意：求一个字符串所有前缀的数量和并mod10007
思路：利用Next数组：前i个字符所组成的字符串的最大前后缀匹配长度的性质，dp[i]=dp[Next[i-1]]+1，dp[i]是以i结尾的字符串数量，由Next可知此字符串也一定是前缀，+1即本身，所有前缀和等于所有的dp[i]相加，总之比较难理解。

#include
#include
#include
#define N 200005
using namespace std;
int Next[N],dp[N];//以i结尾的前缀总数
char s[N];
void getnext(int n)
{
    int i,j;
    for(i=1,j=0; iwhile(s[j]!=s[i]&&j>0)
            j=Next[j-1];
        if(s[i]==s[j])
            j++;
        Next[i]=j;
    }
}
int main()
{
    int t,n,i;
    cin>>t;
    while(t--)
    {
        cin>>n;
        for(i=1; i<=n; i++)
            dp[i]=0;
        scanf("%s",s);
        getnext(n);
        int num=0;
        for(i=1; i<=n; i++)
        {
            dp[i]=dp[Next[i-1]]+1;
            num=(num+dp[i])%10007;
        }
        printf("%d\n",num);
    }
}