【hdu 3336】Count the string（kmp）

Count the string

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 9230    Accepted Submission(s): 4292

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

 

【题解】【kmp】

【kmp的水题，只要将函数建出来，然后将以逐位为结尾的字符串在原串中出现的次数加起来即可，注意多组数据】

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
char s[200010];
int len,t,nxt[200010],ans;
inline void next()
{
    int i,j;
    nxt[0]=-1;
    for (i=0;i<len;++i)
      {
          j=nxt[i];
          while (j!=-1&&s[j]!=s[i]) j=nxt[j];
          nxt[i+1]=j+1;
      }
    return;
}
int main()
{
    int i,j;
    scanf("%d",&t);
    for (i=1;i<=t;++i)
     {
         int k;
        ans=0;
        scanf("%d",&len);
         scanf("%s",s);
         next();
         for (j=1;j<=len;++j)
           {
               k=nxt[j];
               while (k!=-1)
                {k=nxt[k]; ans=(ans+1)%10007;}
           }
        printf("%d\n",ans);
     }
    return 0;
}