1050 String Subtraction （20 分） map

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最新推荐文章于 2023-11-27 12:25:22 发布

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本文介绍了一种计算两个字符串差集的高效算法，通过使用map记录第二个字符串中的字符，然后遍历第一个字符串，去除所有出现在第二个字符串中的字符。该算法适用于字符串长度不超过10^4的情况。

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

给两个字符串s1,s2求s1去掉s2中的字符后的字符串。

思路：

用map记下s2中的字符，然后再对s1中的字符s2是否存在。

输入用的cin.getline();

代码如下：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <map>
#include <string>
using namespace std;
const int maxn=1e4+5;
char s1[maxn],s2[maxn];
map<char,int>ma;
int main()
{
    cin.getline(s1,maxn);
    cin.getline(s2,maxn);
    int len1=strlen(s1),len2=strlen(s2);
    for (int i=0;i<len2;i++)
        ma[s2[i]]=1;
    for (int i=0;i<len1;i++)
    {
        if(!ma[s1[i]])
            printf("%c",s1[i]);
    }
    printf("\n");
    return 0;
}