题目描述
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.
输入描述
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.
输出描述
For each test case, print S1−S2 in one line.
输入例子
They are students.
aeiou
输出例子
Thy r stdnts.
我的代码
#include<stdio.h>
#include<string.h>
int main(){
char a[100055],b[100055];
int hashTable[201]={0};
int len1,len2;
gets(a); //读取有空格的字符串时使用gets()
gets(b);
len1=strlen(a); //不要再循环里面再计算长度
len2=strlen(b);
for(int i=0;i<len2;i++){
hashTable[b[i]]=1; //字符可以自动转换为ASC码
}
for(int i=0;i<len1;i++){
if(hashTable[a[i]]==0){
printf("%c",a[i]);
}
}
return 0;
}
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