1050. String Subtraction (20)

时间限制

10 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.

这道题暴力比较的复杂度时n^2，实不可取，应该用空间换时间，利用一个大小为256的数组exist，首先遍历s2，让s2中出现的字符k对应的ascii码在数组中标记为1，即exist【k】 = 1，然后遍历s1，每个字符k,若exist【k】==1则不输出，否则输出，这样复杂度为n，可以过。

#include <stdio.h>#include <string.h>int main(){int exist[256], len, i;char s2[10001], s1[10001];memset(exist, 0, sizeof(int)*256);gets(s1);gets(s2);len = strlen(s2);for(i=0;i<len;i++){exist[s2[i]] = 1;}len = strlen(s1);for(i=0;i<len;i++){if(exist[s1[i]] == 0)printf("%c",s1[i]);}printf("\n");return 0;}