1050. String Subtraction (20)

题目如下：

Given two strings S₁ and S₂, S = S₁ - S₂ is defined to be the remaining string after taking all the characters in S₂ from S₁. Your task is simply to calculate S₁ - S₂ 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 S₁ and 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 - S₂ in one line.

Sample Input:

They are students.
aeiou

Sample 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 == '
') 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;
}