hdu 3336 KMP之杭电next灵魂

Count the string

Time Limit: 2000/1000 MS (Java/Others)Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1803Accepted Submission(s): 851

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

14abab
sample output
6

/*本题题意可以理解为 把串a的各个子串分别作为模式串 查找模式串在主串中出现的次数
我们可以把串a的next函数打印出来 然后对应子串的next也就是next函数中的一部分
所以就不用一个一个子串的去求子串的next函数了  
另外 我们可以发现当某个子串只在a中出现一次的时候 再往子串上加字符 即下个子串也一定是1
如aabcde  aabc在主串中出现一次 那aabcd肯定更加是1次
所以之后的全是1 就没必要继续调用KMP了 就可以节省很多时间 防止超时
*/
#include<stdio.h>
#include<string.h>
char  a[200005];
char  b[200005];
int next[200005],d,d2,cnt;
void get_next()
{
     int i=1,j=0;
  next[1]=0;
  while(i<=d)
  {
   if(j==0||a[i]==a[j]) {i++;j++;next[i]=j;}
   else j=next[j];
  }
}
int KMP(char *str)
{
      int i=1,j=1;
    while(i<=d)
    {
     if(j==0||b[j]==a[i]) 
     {
      i++;j++;
      if(j==d2+1)  {cnt++; j=1;}
     } 
     else j=next[j];
    }
    return cnt;
}
int main()
{
 int t,i,ans,k;
 scanf("%d",&t);
 while(t--)
 {
  ans=0;
  scanf("%d",&d);
  getchar();
         gets(a+1);
   get_next();
 //  for(i=1;i<=d;i++)
  //  printf("%d ",next[i+1]-1);
 //  printf("\n");
   d2=0;
        for(i=1;i<=d;i++)
  {
   cnt=0;
   b[i]=a[i];
   b[i+1]='\0';
   d2++;
   k=KMP(&b[1]);
  // printf("%d ",k);
   if(ans>10007) ans=ans%10007;
   else
   ans=ans+k;
   if(k==1)
   {
                   ans=(ans+(d-i))%10007;
       break;
   }
  }
     printf("%d\n",ans);
 }
 return 0;
}