1050 String Subtraction (20 分)   Given two strings S​1​​ and S​2​​, S=S​1​​−S​2​​ is defined to be the remaining string after taking all the characters in S​2​​ from S​1​​. Your task is simply to calculate S​1​​−S​2​​ for any given strings. However,…

Given two strings S​1​​ and S​2​​, S=S​1​​−S​2​​ is defined to be the remaining string after taking all the characters in S​2​​ from S​1​​. Your task is simply to calculate S​1​​−S​2​​ for any given strings. However, it might not be that simple to do…

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]]不为true,则输出. 代码如下: #include<iostream> #include<string> using namespace std; bool Hash[128] = {0}; int main(){ string s1, s2; getline(cin, s1);…

1050 String Subtraction Given two strings S​1​​ and S​2​​, S=S​1​​−S​2​​ is defined to be the remaining string after taking all the characters in S​2​​ from S​1​​. Your task is simply to calculate S​1​​−S​2​​ for any given strings. However, it might…

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; ],s2[]; ]; int main(){ ios::sync_with_stdio(false); cin.tie(NULL);…

https://pintia.cn/problem-sets/994805342720868352/problems/994805429018673152 Given two strings S~1~ and S~2~, S = S~1~ - S~2~ is defined to be the remaining string after taking all the characters in S~2~ from S~1~. Your task is simply to calculate S…

题目 Given two strings S1 and S2, S = S1 – S2 is defined to be the remaining string afer 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…

Given two strings S​1​​ and S​2​​, S=S​1​​−S​2​​ is defined to be the remaining string after taking all the characters in S​2​​ from S​1​​. Your task is simply to calculate S​1​​−S​2​​ for any given strings. However, it might not be that simple to do…

1050. String Subtraction (20) 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…

1035 Password (20 分) To prepare for PAT, the judge sometimes has to generate random passwords for the users. The problem is that there are always some confusing passwords since it is hard to distinguish 1 (one) from l (L in lowercase), or 0 (zero) fr…

1073 Scientific Notation (20 分)   Scientific notation is the way that scientists easily handle very large numbers or very small numbers. The notation matches the regular expression [+-][1-9].[0-9]+E[+-][0-9]+ which means that the integer portion has…

1046 Shortest Distance (20 分)   The task is really simple: given N exits on a highway which forms a simple cycle, you are supposed to tell the shortest distance between any pair of exits. Input Specification: Each input file contains one test case. F…

1042 Shuffling Machine (20 分)   Shuffling is a procedure used to randomize a deck of playing cards. Because standard shuffling techniques are seen as weak, and in order to avoid "inside jobs" where employees collaborate with gamblers by performi…

1041 Be Unique (20 分)   Being unique is so important to people on Mars that even their lottery is designed in a unique way. The rule of winning is simple: one bets on a number chosen from [1]. The first one who bets on a unique number wins. For examp…

1015 Reversible Primes (20 分)   A reversible prime in any number system is a prime whose "reverse" in that number system is also a prime. For example in the decimal system 73 is a reversible prime because its reverse 37 is also a prime. Now give…

1152 Google Recruitment (20分) In July 2004, Google posted on a giant billboard along Highway 101 in Silicon Valley (shown in the picture below) for recruitment. The content is super-simple, a URL consisting of the first 10-digit prime found in consec…

简单题. #include<iostream> #include<cstring> #include<cmath> #include<algorithm> #include<cstdio> #include<map> #include<queue> #include<string> #include<vector> using namespace std; +; char t[maxn]; char…

#include <iostream> #include <cstdio> #include <string.h> #include <algorithm> using namespace std; /* 水题,注意字符范围是整个ASCII编码即可. */ ; int vis[maxn]; +]; +]; int main() { gets(s1); //getchar(); gets(s2); int len1=strlen(s1); int len2=s…

this problem  is from PAT, which website is http://pat.zju.edu.cn/contests/pat-a-practise/1050. firstly i think i can use double circulation to solve it ,however the result of two examples is proofed to be running time out. So as the problem said, it…

题意: 给出四组字符串,前两串中第一个位置相同且大小相等的大写字母(A~G)代表了周几,前两串中第二个位置相同且大小相等的大写字母或者数字(0~9,A~N)代表了几点,后两串中第一个位置相同且大小相等的字母所在的位置代表了几分.依照题意输出日期和时间. AAAAAccepted code: #define HAVE_STRUCT_TIMESPEC #include<bits/stdc++.h> using namespace std; string s1,s2,s3,s4; int main(…

The basic task is simple: given N real numbers, you are supposed to calculate their average. But what makes it complicated is that some of the input numbers might not be legal. A legal input is a real number in [−1000,1000] and is accurate up to no m…

一.技术总结 这个是使用了一个bool类型的数组来判断该字符是否应该被输出. 然后就是如果在str2中出现那么就判断为false,被消除不被输出. 遍历str1如果字符位true则输出该字符. 还有需要注意的是memset函数是在头文件#include"cstring"中. 二.参考代码: #include<iostream> #include<cstring> using namespace std; bool hashTable[256]; int main…

Given two strings S​1​​ and S​2​​, S=S​1​​−S​2​​ is defined to be the remaining string after taking all the characters in S​2​​ from S​1​​. Your task is simply to calculate S​1​​−S​2​​ for any given strings. However, it might not be that simple to do…

题意: 输入一个正整数N(<=100),接着输入N行字符串.输出N行字符串的最长公共后缀,否则输出nai. AAAAAccepted code: #include<bits/stdc++.h> using namespace std; ]; ]; ]; int main(){ ios::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL); int n; cin>>n; cin.ignore(); ; ;i<=n;+…

Scientific notation is the way that scientists easily handle very large numbers or very small numbers. The notation matches the regular expression [+-][1-9].[0-9]+E[+-][0-9]+ which means that the integer portion has exactly one digit, there is at lea…

Look-and-say sequence is a sequence of integers as the following: D, D1, D111, D113, D11231, D112213111, ... where D is in [0, 9] except 1. The (n+1)st number is a kind of description of the nth number. For example, the 2nd number means that there is…

1008 Elevator 题目: The highest building in our city has only one elevator. A request list is made up with N positive numbers. The numbers denote at which floors the elevator will stop, in specified order. It costs 6 seconds to move the elevator up one…

题意: 输入一个正整数N(<=1000),接着输入N行数据,每行包括一个ID和一个密码,长度不超过10的字符串,如果有歧义字符就将其修改.输出修改过多少组密码并按输入顺序输出ID和修改后的密码,如果没有修改过就输出There are N accounts and no account is modified.(根据样例如果N==1,are改成is并且accounts不加s) AAAAAccepted code: #define HAVE_STRUCT_TIMESPEC #include<bit…

题意: 电梯初始状态停在第0层,给出电梯要接人的层数和层序号,计算接到所有人需要的时间,接完人后电梯无需回到1层(1层不是0层).电梯上升一层需要6秒,下降一层需要4秒,接人停留时间为5秒. AAAAAccepted code: #include<bits/stdc++.h> using namespace std; ]; int main(){ int n; cin>>n; ; ;i<=n;++i){ cin>>a[i]; ]) ans+=(a[i]-a[i-]…

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 calc…