浙江财经大学
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PAT(A) 1050. String Subtraction (20)

本文由 Ocrosoft 于 2016-12-26 21:54:20 发表

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.
#include <set>
#include <map>
#include <list>
#include <cmath>
#include <stack>
#include <queue>
#include <ctime>
#include <string>
#include <cstdio>
#include <vector>
#include <cctype>
#include <climits>
#include <sstream>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#include <functional>
#define strend string::npos
#define ms(a) memset(a,0,sizeof(a))
#define rep(a,v,b) for(int a=v;a<b;a++)
#define repe(a,v,b) for(int a=v;a<=b;a++)
#define pre(a,v,b) for(int a=v;a>b;a--)
#define pree(a,v,b) for(int a=v;a>=b;a--)
#define lowbit(x) x&-x
typedef long long LL;
const LL LINF = LLONG_MAX / 2;
const int INF = INT_MAX / 2;
const int MAXN = 1e4 + 10;
const int MOD = 1000000007;
int gcd(int a, int b)
{
	if (!b)return a;
	return gcd(b, a%b);
}
/*(◕‿‿◕)(◕‿‿◕) (◕‿‿◕) (◕‿‿◕) (◕‿‿◕) (◕‿‿◕)*/
/*(◕‿‿◕) 签订契约,成为马猴烧酒吧 (◕‿‿◕)*/
/*(◕‿‿◕)(◕‿‿◕) (◕‿‿◕) (◕‿‿◕) (◕‿‿◕) (◕‿‿◕)*/
using namespace std;
int main()
{
	string a, b;
	getline(cin, a);
	getline(cin, b);
	bool os[MAXN] = { false };
	for (int i = 0, len = b.length(); i < len; i++)
		os[b[i]] = true;
	for (int i = 0, len = a.length(); i < len; i++)
		if (!os[a[i]])printf("%c", a[i]);
	return 0;
}

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