【PAT（甲级）】1050 String Subtraction（用map＜char,int＞标记字符）

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.

解题思路：

给出两行字符串，要求在第一行字符串中删去第二行字符串包含的字符。

因为有速度上的要求，所以用两个循环来做肯定是不行的。所以我们在读入第二行的字符时需要对他们做一个标记，带有这个标记的字符在输出的时候就可以不输出，时间复杂度就是O（n）啦；用map就可以很好的解决这个问题。

因为是对每个字符进行标记所以要用char，map这种，用string反而会麻烦。

char读取一行的代码如下：

char b;
b = getchar();//读取字符
while(b != '\n'){//当不是换行的时候
	S1[j]=b;//塞入char数组
	j++;//下标往后移动
    b = getchar();//继续读取下一个字符
}

代码：

#include
using namespace std;
int main(){
	map<char,int> print;//标记该字符是否需要输出
	char S1[10001];//记录S1字符串
	int j=0;//统计S1字符串有几个字符
    char b;
    b = getchar();
	while(b != '\n'){
		S1[j]=b;
		j++;
        b = getchar();
	}
    b = getchar();
	while(b != '\n'){
		print[b]=1;
        b = getchar();
	}
	for(int i=0;i
		if(print.count(S1[i]) > 0 ){
		}
        else{
            cout<
        }
	}
	return 0;
}