字元B中含有多少字元A C++||JAVA
阿新 • • 發佈：2018-12-23
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.
- 樣例輸入
-
3 11 1001110110 101 110010010010001 1010 110100010101011
- 樣例輸出
-
3 0
3
-
import java.util.Scanner; /*for(int i=0;i<n.length()-m.length()+1;i++){ if(m.equals(n.substring(i,i+m.length()))){ ans++; } }*/ public class Main { public static void main(String[] args) { Scanner in=new Scanner(System.in); int s=in.nextInt(); while(s-->0){ int count=0; String a=in.next(); String b=in.next(); for(int i=0;i<b.length()-a.length()+1;i++){ if(a.equals(b.substring(i,i+a.length())))count++; } System.out.println(count); } } }#include<iostream> #include<string.h> #include<algorithm> using namespace std; void ff(); int main(void){ int n; cin>>n; while(n--) ff(); return 0; } void ff(){ int sum=0; string a,b; cin>>a>>b; for(int i=0;i<b.length()-a.length()+1;i++) if(b.compare(i,a.length(),a)==0) sum++; cout<<sum<<endl; }