PAT (Advanced Level) Practice 1050

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最新推荐文章于 2022-07-04 21:31:16 发布

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分类专栏： PAT_甲文章标签： PAT PAT1050 PAT 1050

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本文介绍了一个简单的字符串处理问题，即如何快速计算两个字符串S1和S2的差集S=S1-S2，即将S2中所有字符从S1中去除后的剩余部分。文章通过示例输入输出展示了算法的具体实现，并提供了一段C++代码实现。

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1050 String Subtraction (20)（20 point(s)）

Given two strings S~1~ and S~2~, S = S~1~ - S~2~ is defined to be the remaining string after taking all the characters in S~2~ from S~1~. Your task is simply to calculate S~1~ - S~2~ 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 S~1~ and S~2~, respectively. The string lengths of both strings are no more than 10^4^. 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 S~1~ - S~2~ in one line.

Sample Input:

They are students.
aeiou

Sample Output:

Thy r stdnts.

题意：就是将第一行字符串中删掉第二行出现过的字符，最后将结果字符输出即可。

code

#include<cstdio>
#include<iostream>
#include<cstring>
#include<map>
using namespace std;
#define N 10010
char s1[N], s2[N];
map<char, int> mmap;

int main(){
	cin.getline(s1, sizeof(s1));
	cin.getline(s2, sizeof(s2));
	int len1 = strlen(s1), len2 = strlen(s2);
	for(int i = 0;i < len2;i++){
		mmap[s2[i]]++;
	}
	for(int i = 0;i < len1;i++){
		if(mmap[s1[i]] == 0)
			printf("%c", s1[i]);
	}
	return 0;
}