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
-
3 11 1001110110 101 110010010010001 1010 110100010101011
样例输出 -
3 0 3
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();
}
}