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,…

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…

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);…

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…

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…

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…

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…

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…

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);…

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…

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…

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, it…

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, it…

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,…

PAT甲级:1066 Root of AVL Tree (25分) 题干 An AVL tree is a self-balancing binary search tree. In an AVL tree, the heights of the two child subtrees of any node differ by at most one; if at any time they differ by more than one, rebalancing is done to rest…

PAT甲级:1064 Complete Binary Search Tree (30分) 题干 A Binary Search Tree (BST) is recursively defined as a binary tree which has the following properties: The left subtree of a node contains only nodes with keys less than the node's key. The right subtre…

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…

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…

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;+…

题意:给出两个字符串s1和s2,在s1中删去s2中含有的字符. 思路:注意,因为读入的字符串可能有空格,因此用C++的getline(cin,str).PAT系统迁移之后C语言中的gets()函数被禁用了. 代码: #include <iostream> #include <string> using namespace std; int main() { string str1,str2; getline(cin,str1); getline(cin,str2); ]={fals…

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 Specific…

题意: 输入三个正整数N,M,S(N<=100,M<N,S<=2^30)分别代表数的结点个数,非叶子结点个数和需要查询的值,接下来输入N个正整数(<1000)代表每个结点的权重,接下来输入M行,每行包括一个两位数字组成的数代表非叶子结点的编号以及数字x表示它的孩子结点个数,接着输入x个数字表示孩子结点的编号.以非递增序输出从根到叶子结点的路径权重,它们的和等于S. AAAAAccepted code: #define HAVE_STRUCT_TIMESPEC #include<…

题意: 输入两个正整数N和K(N<=40000,K<=2500),分别为学生和课程的数量.接下来输入K门课的信息,先输入每门课的ID再输入有多少学生选了这门课,接下来输入学生们的ID.最后N次询问,输入学生的ID输出该学生选了多少们课,输出所选课程的数量,按照递增序输出课程的ID. trick: 第5个数据点出现段错误,把原本以map存学生对应ID再映射vector存储该学生所选课程改成vector嵌套在map内,就没有段错误的问题出现,疑似映射过程中指针漂移??? AAAAAccepted…

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…

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…