pat1050

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.

题目含义是要找出一个不包含一些部分字符的字符串，问题关键在于怎么判断字符串2中的这个字符是否在字符串1中：

可以设计一个不存在列表，里面包含了所有的字符，值是true&false，扫描一遍字符串2，确定哪些要改成true的放在不存在列表中，等到扫描字符串1的时候就看这个字符是否为true，如果不存在，就不输出

#include 
#include
using namespace std;

int main(int argc, const char * argv[]) {
    char s1[10002];
    char s2[10002];
    bool bucunzai[300]={false};
    cin.getline(s1,10002);
    cin.getline(s2,10002);
    int l2=strlen(s2);
    for(int i=0;i