1050 String Subtraction (20)（20 point(s)）
Given two strings S~1~ and S~2~, S = S~1~ - S~2~ is defined to be the remaining string after taking all the characters in S~2~ from S~1~. Your task is simply to calculate S~1~ - S~2~ 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 S~1~ and S~2~, respectively. The string lengths of both strings are no more than 10^4^. 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 S~1~ - S~2~ in one line.
Sample Input:
They are students.
aeiou
Sample Output:
Thy r stdnts.
题意：就是将第一行字符串中删掉第二行出现过的字符，最后将结果字符输出即可。
code
#include<cstdio>
#include<iostream>
#include<cstring>
#include<map>
using namespace std;
#define N 10010
char s1[N], s2[N];
map<char, int> mmap;
int main(){
cin.getline(s1, sizeof(s1));
cin.getline(s2, sizeof(s2));
int len1 = strlen(s1), len2 = strlen(s2);
for(int i = 0;i < len2;i++){
mmap[s2[i]]++;
}
for(int i = 0;i < len1;i++){
if(mmap[s1[i]] == 0)
printf("%c", s1[i]);
}
return 0;
}
