5 Binary String Matching【kmp】

Binary String Matching

时间限制：3000 ms  |  内存限制：65535 KB

难度：3

描述

Given two 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’ while the pattern string A is ‘11’, you should output 3, because the pattern A appeared at the posit

输入

The first line consist only one integer N, indicates N cases follows. In each case, there are two lines, the first line gives the string A, length (A) <= 10, and the second line gives the string B, length (B) <= 1000. And it is guaranteed that B is always longer than A.

输出

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

样例输入

31110011101101011100100100100011010110100010101011 

样例输出

303 

查找一个字符串在另外一个字符串中出现的次数，kmp 算法，比较简单的就可以做出来，kmp虽然理解比较难理解，但是要记住它的运行流程和代码实现还是比较简单的，

如果实在是理解不了，可以先记住，以后学的东西多了，慢慢对很多的算法的理解就深入了，也许就慢慢懂了....

ps：个人目前也不是太懂，只是可以正确的敲出代码，另外c++里面的数据类型，string 里面有查找函数，可以直接解决这类问题....

#include<stdio.h>#include<string.h>char x[1005],y[10005];int lenx,leny,next[1005];void get_next()//找到特征数组{int i=0,j=-1;next[0]=-1;while(i<lenx){if(j==-1||x[i]==x[j]){++i;++j;next[i]=j;}else{j=next[j];}}}void kmp()//调用函数{get_next();int i=0,j=0,cnt=0;while(i<leny){if(j==-1||x[j]==y[i]){++i;++j;if(j==lenx)//记录出现次数{++cnt;}}else{j=next[j];}}printf("%d\n",cnt);}int main(){int t;scanf("%d",&t);getchar();while(t--){gets(x);gets(y);lenx=strlen(x);leny=strlen(y);//长度比较好用kmp();} }