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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
来源
网络
#include<algorithm> #include<iostream> #include<string> #include<cstring> using namespace std; int main() { int n;cin>>n;string s,s1,s2; while(n--) { string::size_type position=0; //position = s.find("you"); //find 函数 返回you 在s 中的下标位置 //cout<<flag<<endl; cin>>s1>>s2;int cont =0;while((position=s2.find(s1,position))!=(string::npos)) //find 必须匹配完整的字符串,find_first_of只需要匹配部分 { cont++;position++; } cout<<cont<<endl; } return 0; }
或者:
#include<iostream> #include<string> using namespace std; int main() {int n;cin>>n;while(n--){ int count=0;string str1,str2,a="";cin>>str1>>str2;for(int i=0;i<str2.size();i++){a=str2.substr(i,str1.size()); //substr() 从一个字符串复制一个从指定位置开始,并具有指定长度的子字符串。里边两个参数,第一个起始位置,第二个提取字符数if(a==str1) count++;}cout<<count<<endl;} }
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