1050 String Subtraction （20 分)

字符串减法算法

最新推荐文章于 2024-05-29 11:50:32 发布

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最新推荐文章于 2024-05-29 11:50:32 发布

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本文介绍了一个简单的字符串减法算法问题，即从一个字符串中移除另一个字符串中的所有字符。通过使用128大小的数组来标记ASCII字符是否需要被删除，并利用getline函数正确处理输入中的空格与换行。

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.

解题思路：1.ASCII码工128个十进制由0-127 因此用一个128大小的数组存足以

2.注意输入函数：getline(cin,str)可以接收空格，换行符结束；而普通cin和getline都不能接受空格

3.如果使用erase函数，i会改变，不妥

#include <iostream>
#include <bits/stdc++.h>
using namespace std;

int vis[128];//ASCII码工128个字符 十进制0到127

int main()
{
    string s1,s2,ans;
    getline(cin,s1);
    getline(cin,s2);
    memset(vis,0,sizeof(vis));
    for(int i = 0; i < s2.size(); i++){
        vis[s2[i]] = 1;
    }
    for(int i = 0; i < s1.size(); i++){
        if(!vis[s1[i]]){
            ans += s1[i];
        }
    }
    cout << ans;
    return 0;
}