Nyoj5 Binary String Matching (KMP)

Binary String matching time limit: theMs | Memory Limit:65535KB Difficulty:3

Describe

Given strings A and B, whose alphabet consist only ' 0 ' and ' 1 '. Your task is-only to-tell how many times does A appear as a substring of B? For example, the text string B is ' 1001110110 ' and the pattern string A is ' one ', you should output 3, because the patter N A appeared at the posit

Input

The first line consist only one integer n, indicates n cases follows. In each case, there is lines, the first line gives the string A, Length (a) <=, and the second line gives the S Tring B, Length (b) <= 1000. And it is guaranteed this B is always longer than A.

Output

For each case, the output a single line consist a single integer, tells how the many times do B appears as a substring of a.

Sample input

Sample output

Test instructions: Find out the number of occurrences of the pattern string in the main string.

Naive algorithm Code:

#include <cstdio> #include <cstring>int main () {  int n,count;  Char a[200],b[1200];  scanf ("%d", &n);  GetChar ();  while (n--)  {    count=0;    int I=0,j=0,len;    scanf ("%s\n%s", b);    Len=strlen (b);    while (I<=len)    {      if (a[j]== ')      } {        count++;        i=i-j+1;         j=0;      }      else if (A[j]==b[i])      {        i++;        j + +;      }      else      {        i=i-j+1;//The key is backtracking        j=0;      }    }    printf ("%d\n", count);  }  return 0;}

KMP algorithm:

#include <cstdio> #include <cstring>int nextval[200];void get_next (char a[])//Get next array; {  int len;  int i=0,j=-1;  Nextval[0]=-1;  Len=strlen (a);  while (I<=len)  {    if (J==-1 | | a[i]==a[j])    {      ++i;      ++j;      if (A[i]==a[j])      nextval[i] = nextval[j];  Replace the backtracking content with the next array;      else      nextval[i] = j;    }    else      j=nextval[j];}  } int KMP (char a[],char b[])//kmp the body function {  int i=0,j=0,count=0;  int Lena,lenb;  Lena=strlen (a);  Lenb=strlen (b);  Get_next (a);  while (I<=lenb)  {    if (J==-1 | | a[j]==b[i])    {      ++i;      ++j;    }    else      J=nextval[j];    if (J>=lena)    {      count++;      J=NEXTVAL[J];    }  }  return count;} int main () {  int n;  Char a[20],b[1200];  scanf ("%d", &n);  while (n--)  {    scanf ("%s\n%s", A, b);    printf ("%d\n", KMP (A, b));  }  return 0;}

Nyoj5 Binary String Matching (KMP)