题目
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:
代码语言：javascript复制They are students.
aeiouSample Output:
代码语言：javascript复制Thy r stdnts.解析
- 所谓
s1 - s2就是从s1中去除出现在s2中的字符，输出剩余部分。 - 所以可以逐个判断
s1的字符，判断其是否在s2中出现，若未出现则输出，否则跳过即可。 - 为了操作简单，使用一个布尔数组
flag[256]，ascii字符也就是0-255，所以遍历一次s2，将flag[s2[i]]设置为true即表示当前字符在s2中出现过。
代码
代码语言：javascript复制#include <iostream>
#include <string>
#include <string.h>
using namespace std;
int main() {
string s1, s2;
// 原字符串中包含空格，使用getline
getline(cin, s1);
getline(cin, s2);
int len1 = s1.length(), len2 = s2.length();
bool flag[256] = {false};
// 在s2中出现的字符，在s1中剔除
for (int i = 0; i < len2; i) flag[s2[i]] = true;
// 逐个字符判断，若未在s2中出现则输出
for (int i = 0; i < len1; i)
if (!flag[s1[i]]) cout << s1[i];
return 0;
}
