Binary String Matching
2018-03-21 20:49
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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
#include<iostream> #include<string> using namespace std; int main() { int N; cin>>N; while(N--) { string s,a,b; cin>>a>>b; int at =0,i=0,len; while((i=b.find(a,i))!=(string::npos)) { at++; i++; } cout<<at<<endl; } return 0; }
注解: i=b.find(a,i) //在字符串b中查找字符串a从b字符中第i个元素开始查找,返回为int型的数值重新赋值给i,string::npos是标准库的string容器属性。返回字符存放位置。 这个东西是一个容器,它将字符串分成一个一个来存储。 while((i=b.find(a,i))!=(string::npos)) { at++; //计算器 i++; } //对b中字符串逐次与a比较直到结束
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