EASY_ZJU_PAT_ADVANCED LEVEL 1050 高效輸入含有空格的字符串 高效刪除字符串中的字符

1050. String Subtraction (20)

時間限制

10 ms

內存限制

32000 kB

代碼長度限制

16000 B

判題程序

Standard

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.

NOTE：

1.通常的cin和scanf("%s",*char)這種遇到空格會停止，有系統內置的函數可以實現連續帶空格的輸入，gets(char*)和getline(cin,string)，但是這兩種的效率都不是很高。可以用

	while(scanf("%c",*char)&&ch!='\n'){

	}

	這種形式進行帶有空格的字符串輸入

2.這道題思路不難，但是重點就在如何減少運行時間。要不然會出現超時的情況。

/************************************************
	@ AUTHOR	: GAOMINQUAN
	@ DATA		: 2014 - 2 - 24
	@ MAIL		: [email protected]
	@ HARD		: EASY ***

/************************************************/
#include<iostream>
#include<vector>
#include<string.h>
using namespace std;
const int M = 10000;
const int charNum = 127;
int main(){
	char line1[M];
	vector<int> charMark(charNum,0);
	char ch;
	int index1 = 0;

	while(scanf("%c",&ch)&&ch!='\n'){
		line1[index1++] = ch;
	}
	while(scanf("%c",&ch)&&ch!='\n'){
		int index = ch;
		charMark[index] = -1;
	}

	for(int line1Index = 0; line1Index<index1; line1Index++){
		int index = line1[line1Index];
		if(charMark[index] == 0){
			printf("%c",line1[line1Index]);
		}
	}
	cout<<endl;
	return 0;

}