Binary String Matching時(shí)間限制：??3000??ms ?|? 內(nèi)存限制：??65535??KB?難度：??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?輸入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.輸出For each case, output a single line consist a single integer, tells how many times do B appears as a substring of A.import?java.util.Scanner;?? ?? public?class?Main?{?? ????public?static?void?main(String[]?args)?{?? ????????Scanner?input?=?new?Scanner(System.in);?? ????????int?N?=?input.nextInt();?? ????????String?s1,s2; ???????? ???????? ????????for(int?i?=0;i<N;i++){ ???????? s1=input.next(); ???????? s2=input.next(); ???????? if(s1.length()>s2.length()||s1.length()>10||s2.length()>1000){ ???????? System.out.println("Wrong!"); ???????? break; ???????? } ???????? ???????? else ???????? System.out.println(count(s1,s2)); ????????} ????} ????public?static?int?count(String?s1,String?s2){ ???? int?num=0; ???? ???? for(int?i=?0;i<s2.length()-1;i++){ ???? for(int?j=?i+1;j<s2.length();j++){ ???? if(s2.substring(i,?j)==s1) ???? num++; ???? } ???? } ???? ???? return?num; ????} }