1050. String Subtraction (20)

最新推荐文章于 2021-10-09 14:13:36 发布

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最新推荐文章于 2021-10-09 14:13:36 发布

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分类专栏： PAT习题

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本文介绍了一个简单的字符串减法问题：给定两个字符串S1和S2，计算S1-S2的结果，即从S1中去除所有S2中存在的字符。通过使用哈希映射的方式，快速实现了这一计算过程。

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1050. String Subtraction (20)

时间限制

100 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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.

水题，利用hash映射可以解决，本以为char与int的转换需要写个函数，没想到可以直接用。

#include<iostream>
#include<fstream>
#include<string>
#include<algorithm>
#include<vector>
using namespace std;

int main()
{
	
	//ifstream fin("Text.txt");
	string s1, s2;
	char c;
	getline(cin, s1);
	getline(cin, s2);
	int *asc = new int[200]{ 0 };
	int i, j;
	vector<int>order;
	for (i = 0; i < s1.size(); i++)
	{
		c = s1[i];
		//j = str2num(c);
		asc[c] = 1;
		order.push_back(c);
	}
	for (i = 0; i < s2.size(); i++)
	{
		c = s2[i];
		//j = str2num(c);
		asc[c] = 0;
	}
	string res;
	for (i = 0; i < order.size(); i++)
	{
		if (asc[order[i]] == 1)res.push_back(order[i]);
	}
	cout << res<<endl;
	//cin.get();
	return 0;
}