ACM Binary String Matching

Binary String Matching

時間限制： 3000  ms  |  內存限制： 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.

樣例輸入

3
11
1001110110
101
110010010010001
1010
110100010101011 

樣例輸出

3
0
3 


程序代碼:

 1 #include<iostream> 
 2 #include<string> 
 3 using namespace std; 
 4 int main() { 
 5     int N;
 6     cin>>N;
 7     while(N--) {
 8         string s,a,b;
 9         cin>>a>>b;
10         int at =0,i=0,len;
11         while((i=b.find(a,i))!=(string::npos)) { 
12             at++;
13             i++; 
14         } 
15         cout<<at<<endl; 
16     } 
17 }

註解:
i=b.find(a,i)   //在字符串b中查找字符串a從b字符中第i個元素開始查找,返回爲int型的數值從新賦值給i,string::npos是標準庫的string容器屬性。返回字符存放位置。 這個東西是一個容器，它將字符串分紅一個一個來存儲。while((i=b.find(a,i))!=(string::npos)) {    at++;   //計算器   i++;       }  //對b中字符串逐次與a比較直到結束