【PAT甲级】1050 String Subtraction (20 分)
题意:
输入两个串,长度小于10000,输出第一个串去掉第二个串含有的字符的余串。
trick:
ascii码为0的是NULL,减去'0','a','A',均会导致可能减成负数。
AAAAAccepted code:
#define HAVE_STRUCT_TIMESPEC
#include<bits/stdc++.h>
using namespace std;
char s1[],s2[];
int vis[];
int main(){
ios::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
cin.getline(s1,);
cin.getline(s2,);
int n=strlen(s1);
int m=strlen(s2);
for(int i=;i<m;++i)
vis[s2[i]-NULL]=;
for(int i=;i<n;++i)
if(!vis[s1[i]-NULL])
cout<<s1[i];
return ;
}
【PAT甲级】1050 String Subtraction (20 分)的更多相关文章
- PAT 甲级 1050 String Subtraction (20 分) (简单送分,getline(cin,s)的使用)
1050 String Subtraction (20 分) Given two strings S1 and S2, S=S1−S2 is defined to be t ...
- PAT Advanced 1050 String Subtraction (20 分)
Given two strings S1 and S2, S=S1−S2 is defined to be the remaining string after taking ...
- PAT练习--1050 String Subtraction (20 分)
题⽬⼤意:给出两个字符串,在第⼀个字符串中删除第⼆个字符串中出现过的所有字符并输出. 这道题的思路:将哈希表里关于字符串s2的所有字符都置为true,再对s1的每个字符进行判断,若Hash[s1[i] ...
- PAT甲级——1050 String Subtraction
1050 String Subtraction Given two strings S1 and S2, S=S1−S2 is defined to be the remain ...
- PAT 甲级 1050 String Subtraction
https://pintia.cn/problem-sets/994805342720868352/problems/994805429018673152 Given two strings S~1~ ...
- PAT Advanced 1050 String Subtraction (20) [Hash散列]
题目 Given two strings S1 and S2, S = S1 – S2 is defined to be the remaining string afer taking all th ...
- 1050 String Subtraction (20分)
Given two strings S1 and S2, S=S1−S2 is defined to be the remaining string after taking ...
- PAT 解题报告 1050. String Subtraction (20)
1050. String Subtraction (20) Given two strings S1 and S2, S = S1 - S2 is defined to be the remainin ...
- PAT 甲级 1035 Password (20 分)
1035 Password (20 分) To prepare for PAT, the judge sometimes has to generate random passwords for th ...
随机推荐
- codeforces 597div2 F. Daniel and Spring Cleaning(数位dp+二维容斥)
题目链接:https://codeforces.com/contest/1245/problem/F 题意:给定一个区间(L,R),a.b两个数都是属于区间内的数,求满足 a + b = a ^ b ...
- JS高级---复习
复习 面向过程和面向对象都是编程的思想, 方式不一样 面向过程: 凡事都是亲力亲为, 所有的代码都要自己写, 每一步都要很清楚, 注重的是过程 面向对象: 执行者成为指挥者, 只要找对象, 然后让对象 ...
- luckyframe的一些坑
建议使用idea运行 1.第一次运行访问http://localhost:8088/LuckyFrameServer 2.luckyframe提示“javax.net.ssl.SSLKeyExcept ...
- casperJs的安装2
通过上面一节,以为能够顺利安装好phantom 没想到,根本没成功. 接着,通过github上下载项目,重新安装: 1.clone项目:[https://github.com/ariya/phanto ...
- Bugku-CTF之login3(SKCTF)(基于布尔的SQL盲注)
Day41 login3(SKCTF)
- Codeforces Round #622 C2.Skyscrapers (hard version)
This is a harder version of the problem. In this version n≤500000n≤500000 The outskirts of the capit ...
- Reinforcement Learning Algorithm 资源
算法源码: https://github.com/ljpzzz/machinelearning https://github.com/imraviagrawal/Reinforcement-Learn ...
- STA之RC网
STA的主要工作是计算电路网络的延时,如今的电路网络还是由CMOS cell和net组成的,所以STA所要计算的延时仍是电容的充放电时间.等量子计算机普及的时候,如今的这一套理论都将随着科技的进步被丢 ...
- .net core 删除主表,同时删除子表
前提条件: 代码懒加载, 数据库有外键关联 var entity = context.主表.Include(o => o.子表).FirstOrDefault(p => p.Id == i ...
- ClickOnce部署,修改deploymentProvider
打开 以.application 为后缀的文件,找到并修改 <deploymentProvider codebase="" /> 即可