Section 1.2 Name That Number
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一个匹配搜索的问题,不过是我第一次写从文件中即时读取即时搜索。
开始的思路是通过input把number转为name,再进行搜索,后来发现这样不仅费时而且浪费空间;于是采用了把data.txt中的name转化为number,依次与input匹配的方法。
/*ID: alexyua2PROG: namenumLANG: C++*/#include<fstream>#include<string>using namespace std;ifstream fin("namenum.in");ofstream fout("namenum.out");//ifstream fin("in.txt");//ofstream fout("out.txt");ifstream data("dict.txt");int main(){ //get input string num; fin>>num; /*transform every name in dict.txt to number then compare it with the input*/ string name; int i; int state = 0;//no matching while(data>>name) { int len = name.length(); string tempt; for(i=0;i<len;i++) { if(name[i] < 'Q') tempt.push_back((name[i] - 'A')/3 + '2'); else tempt.push_back((name[i] - 'Q')/3 + '7'); } if(tempt == num) { state = 1; fout<<name <<endl; } tempt.clear(); } if(state == 0) fout<<"NONE" <<endl; //close files fin.close(); fout.close(); data.close(); return 0;}
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