L1-011 A-B
本题要求你计算A−B。不过麻烦的是，A和B都是字符串 —— 即从字符串A中把字符串B所包含的字符全删掉，剩下的字符组成的就是字符串A−B。
输入格式：
输入在 2 行中先后给出字符串A和B。两字符串的长度都不超过 104，并且保证每个字符串都是由可见的 ASCII 码和空白字符组成，最后以换行符结束。
输出格式：
在一行中打印出A−B的结果字符串。
输入样例：
tex
I love GPLT! It's a fun game!
aeiouI love GPLT! It's a fun game!
aeiou输出样例：
tex
I lv GPLT! It's fn gm!I lv GPLT! It's fn gm!Solution：
java
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
public class Main {
public static void main(String[] args) throws IOException {
BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
char[] a = in.readLine().toCharArray();
char[] b = in.readLine().toCharArray();
StringBuilder sb = new StringBuilder();
int[] book = new int[128];
for (char c : b) {
book[c] = 1;
}
for (char c : a) {
if (book[c] != 1) sb.append(c);
}
System.out.println(sb);
}
}import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
public class Main {
public static void main(String[] args) throws IOException {
BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
char[] a = in.readLine().toCharArray();
char[] b = in.readLine().toCharArray();
StringBuilder sb = new StringBuilder();
int[] book = new int[128];
for (char c : b) {
book[c] = 1;
}
for (char c : a) {
if (book[c] != 1) sb.append(c);
}
System.out.println(sb);
}
}