1050 String Subtraction (20)

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原文链接：http://www.cnblogs.com/mr-stn/p/9178768.html

本文介绍了一种计算两个字符串差集的简单算法，并提供了一个C++实现示例。该算法通过使用字符映射的方式快速地从第一个字符串中移除第二个字符串中存在的字符。

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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.

注意点：连个字符串都可能包含空格， 要用getline输入

#include<iostream>
#include<vector>
using namespace std;
int main(){
  string s, s1;
  getline(cin, s);
  getline(cin, s1);
  int i;
  vector<int> v(128,-1);
  for(i=0; i<s1.size(); i++) v[s1[i]] = 1;
  for( i=0; i<s.size(); i++)
  if(v[s[i]]!=1) cout<<s[i];
  return 0;
}