1050. String Subtraction (20)

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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，如何快速地计算出S1-S2的结果，即去除S1中所有出现在S2中的字符。输入包括两个字符串，输出为处理后的结果。

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.

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cstring>
#include <queue>
#include <stack>

using namespace std;

int book[10000]={0};
int main()
{
    freopen("in.txt","r",stdin);

    char a[11000];
    char b[11000];
    gets(a);//读入带空格的字符串
    gets(b);
    int len1=strlen(a);//strlen获取字符串长度
    int len2=strlen(b);
    int i;
    cout<<len1<<' '<<len2;
    /*
    for(i=0;i<len2;i++)
    {
        int t;
        t=b[i];
        book[t]=1;
    }
    for(i=0;i<len1;i++)
    {
        int t2;
        t2=a[i];
        if(book[t2]==1)
        {
            continue;
        }
        else
        {
            printf("%c",a[i]);
        }
    }
    */
    return 0;
}