hdu3336 Count the string

地址：http://acm.hdu.edu.cn/showproblem.php?pid=3336

题目：

Count the string

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 9884    Accepted Submission(s): 4617

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

 

 

Recommend

lcy

 思路：next数组+dp

#include <cstdio>
#include <cstring>
#include <iostream>

using namespace std;

#define MP make_pair
#define PB push_back
typedef long long LL;
const double eps=1e-8;
const int K=1e6+7;
const int mod=1e9+7;

int nt[K],ans[K],dp[K];
char sa[K],sb[K];
void kmp_next(char *T,int *next)
{
    next[0]=0;
    for(int i=1,j=0,len=strlen(T);i<len;i++)
    {
        while(j&&T[i]!=T[j]) j=next[j-1];
        if(T[i]==T[j])  j++;
        next[i]=j;
    }
}
int kmp(char *S,char *T,int *next)
{
    int ans=0;
    int ls=strlen(S),lt=strlen(T);
    kmp_next(T,next);
    for(int i=0,j=0;i<ls;i++)
    {
        while(j&&S[i]!=T[j]) j=next[j-1];
        if(S[i]==T[j])  j++;
        if(j==lt)   ans++;
    }
    return ans;
}
int main(void)
{
    int t,n;cin>>t;
    while(t--)
    {
        int ans=0;
        scanf("%d%s",&n,sa);
        kmp_next(sa,nt);
        dp[0]=1;
        for(int i=1;i<n;i++)
            dp[i]=dp[nt[i]-1]+1;
        for(int i=0;i<n;i++)
            ans=(ans+dp[i])%10007;
        printf("%d\n",ans);
    }
    return 0;
}