PAT甲级 -- 1050 String Subtraction (20 分)

 

Given two strings S​1​​ and S​2​​, S=S​1​​−S​2​​ is defined to be the remaining string after taking all the characters in S​2​​ from S​1​​. Your task is simply to calculate S​1​​−S​2​​ for any given strings. However, it might not be that simple to do it fast.

Input Specification:

Each input file contains one test case. Each case consists of two lines which gives S​1​​ and S​2​​, respectively. The string lengths of both strings are no more than 10​4​​. It is guaranteed that all the characters are visible ASCII codes and white space, and a new line character signals the end of a string.

Output Specification:

For each test case, print S​1​​−S​2​​ in one line.

Sample Input:

They are students. aeiou

Sample Output:

Thy r stdnts.

3分：

getchar（）把第一个字符读取了

#include <iostream> #include <string> using namespace std; string s1, s2; bool asc[256] = {false}; int main() { getline(cin,s1); getchar(); getline(cin,s2); int len1 = s1.length(), len2 = s2.length(); for(int i = 0; i < len2; i++) { asc[s2[i]] = true; } for(int i = 0; i < len1; i++) { if(asc[s1[i]] == true) continue; else printf("%c",s1[i]); } return 0; }

 20分：

#include <iostream> #include <string> using namespace std; string s1, s2; bool asc[256] = {false}; int main() { getline(cin,s1); getline(cin,s2); int len1 = s1.length(), len2 = s2.length(); for(int i = 0; i < len2; i++) { asc[s2[i]] = true; } for(int i = 0; i < len1; i++) { if(asc[s1[i]] == true) continue; else printf("%c",s1[i]); } return 0; }