PAT 1050. String Subtraction

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本文介绍了一个简单的字符串减法算法，即从字符串S1中移除所有出现在S2中的字符，并提供了一个C++实现示例。该算法适用于字符串长度不超过10^4的情况。

（1）一行字符串的输入也可以用gets，也可以用getLine

（2）gets要赋值到一个指定了大小的char数组中，赋值到char * 好像会跑出运行时异常

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.

#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

int cnt[256];

int main()
{
    // 没有说明长度是不行的
    //char* cs1;
    //char* cs2;

    char cs1[10005], cs2[10005];

    gets(cs1);
    gets(cs2);

    int len1 = strlen(cs1);
    int len2 = strlen(cs2);
    for(int i=0; i<len2; i++)
        cnt[cs2[i]] = 1;

    for(int i=0; i<len1; i++)
        if(cnt[cs1[i]] == 0)
            printf("%c", cs1[i]);

    return 0;
}