PAT 甲级 1050 String Subtraction (20 分)

最新推荐文章于 2020-09-13 15:17:49 发布

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最新推荐文章于 2020-09-13 15:17:49 发布

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本文介绍了一种高效解决字符串相减问题的方法，通过使用C++的map数据结构，能够快速找出并输出第一个字符串中未出现在第二个字符串中的所有字符。

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1050 String Subtraction (20 分)

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.

思路：

用一个map<char,int>存第二个字符串，如果第一个字符串的字符没有在第二个字符里出现，就输出。

#include<iostream>
#include<cstring>
#include<map>
using namespace std;
int main()
{
	string a,b;
	int i,j;
	map<char,int> c;
	getline(cin,a);
	getline(cin,b);
	for(i=0;i<b.size();i++)
	{
		if(c.count(b[i])==0)
		c.insert(make_pair(b[i],1));
	}
	for(i=0;i<a.size();i++)
	{
		if(c.count(a[i])!=1)
		cout<<a[i];
	}
	cout<<endl;
	return 0;
}