题目如下：
Given two strings S1and S2, S = S1- S2is defined to be the remaining string after taking all the characters in S2from S1. Your task is simply to calculate S1- S2for any given strings. However, it might not be that simple to do itfast.
Input Specification:
Each input file contains one test case. Each case consists of two lines which gives S1and S2, respectively. The string lengths of both strings are no more than 104. 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 S1- S2in one line.
Sample Input:They are students.aeiouSample Output:Thy r stdnts.
题目要求从字符串S1中删除所有S2中有的字符，最简单的解决方法就是S1用string，S2用map，然后遍历输出S1，对于S1中的每一个字符，如果map中有，则不输出。
需要注意的是对空格的处理，使用getline(cin,str)可以读取一行，从而实现string中存储带空格字符串，，使用getchar()函数可以捕捉到空格和回车，以此输入S2到map。
代码如下：
#include <iostream>#include <string>#include <string.h>#include <stdio.h>#include <map>using namespace std;int main(){string input,del;getline(cin,input);map<char,int> strs;while(1){char c = getchar();if(c == '\n') break;strs[c] = 1;}for(int i = 0; i < input.length(); i++){char c = input[i];if(strs.find(c) != strs.end()) continue;printf("%c",c);}cout << endl;return 0;}
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