pat 1050. String Subtraction (20)

1050. String Subtraction (20)

时间限制

10 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.

解：暴力卡时，后来听陈叔叔的建议，预处理了下s2，考虑这种情况，如s2=aaa，其实只是a而已，我们预处理为s2=a，可以在一定程度上降低时间复杂度。事实证明是可行的。

代码：

#include<iostream>
#include<cstdlib>
#include<cstdio>
#include<cstring>
#include<set>
using namespace std;
int main()
{
    char s1[10000+1];
    char s2[10000+1];
    string res;
    set<char> sss;
    gets(s1);
    gets(s2);
    int len1=strlen(s1);
    int len2=strlen(s2);
    for(int i=0;i<len2;i++)   //预处理s2
        sss.insert(s2[i]);
    set<char>::iterator p;
    for(int i=0;i<len1;i++)
    {
        int flag=1;
        for(p=sss.begin();p!=sss.end();p++)
        {
            if(s1[i]==*p)
            {
                flag=0;
                break;
            }
        }
        if(flag==1)
           res.push_back(s1[i]);
    }
    cout<<res<<endl;
}

底下是改进的办法，用了set的内置函数find()：

#include<iostream>
#include<cstdlib>
#include<cstdio>
#include<cstring>
#include<set>
using namespace std;
int main()
{
    char s1[10000+1];
    char s2[10000+1];
    string res;
    set<char> sss;
    gets(s1);
    gets(s2);
    int len1=strlen(s1);
    int len2=strlen(s2);
    for(int i=0;i<len2;i++)   //预处理s2
        sss.insert(s2[i]);
    for(int i=0;i<len1;i++)
    {
        int flag=1;
        if(sss.find(s1[i])!=sss.end())
        {
            flag=0;
        }
        if(flag==1)
           res.push_back(s1[i]);
    }
    cout<<res<<endl;
}

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