PAT (Advanced Level) Practise 1050 String Subtraction (20)

最新推荐文章于 2021-07-03 17:45:46 发布

Wang_128

最新推荐文章于 2021-07-03 17:45:46 发布

阅读量240

点赞数

CC 4.0 BY-SA版权

分类专栏： PAT/PTA甲级 ----水题

本文链接：https://blog.youkuaiyun.com/a664607530/article/details/75195848

----水题同时被 2 个专栏收录

234 篇文章

订阅专栏

PAT/PTA甲级

174 篇文章

订阅专栏

本文介绍了一个简单的字符串处理问题——字符串减法。通过遍历并标记第二个字符串中的字符，在第一个字符串中去除这些被标记的字符，从而实现两字符串相“减”。此算法适用于字符串长度不超过10^4的情况。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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.

题意：给你两个字符串，第一个字符串中删除第二个字符串出现的所有字符

解题思路：把第二个字符串出现的字符进行标记即可

#include <iostream>  
#include <cstdio>  
#include <cmath>  
#include <cstring>  
#include <algorithm>  
#include <stack>  
#include <queue>  
#include <vector>  
#include <map>  
  
using namespace std;  
  
char s1[10009],s2[10009];
int visit[10009];
  
int main()  
{  
    gets(s1);
    gets(s2);
    for(int i=0;s2[i];i++)
        visit[s2[i]]=1;
    for(int i=0;s1[i];i++)
        if(!visit[s1[i]])
            printf("%c",s1[i]);
    return 0;  
}