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…

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,…

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…

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…

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 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…

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);…

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,…

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…

#include<iostream> #include<string.h> #include<stdio.h> using namespace std; #define N 128 int main() { ,sum; bool is_exist[N]; char ch; ]; //void *memset(void *s,int ch,size_t n) //将s所指向的某一块内存中的前n个字节的内容全部设置为ch指定的ASCII值 memset(is_exist,,…

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…

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 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…

Source: PAT A1050 String Subtraction (20 分) Description: 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…

String的20个方法 面试题 1.new和不new的区别 String A="OK"; String B="OK";//会去常量池查找有没有"Ok"这个常量,没有就在常量池创建 String C=new String("OK"); String D=new String("OK");//每次new都是创建一个新地址 不new涉及到内存常量池查找机制 2.String.StringBuffer.Strin…

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 Sp…

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 Sp…

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) 时间限制 10 ms 内存限制 32000 kB 代码长度限制 16000 B 判题程序 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…

博主欢迎转载,但请给出本文链接,我尊重你,你尊重我,谢谢~http://www.cnblogs.com/chenxiwenruo/p/6102219.html特别不喜欢那些随便转载别人的原创文章又不给出链接的所以不准偷偷复制博主的博客噢~~ 时隔两年,又开始刷题啦,这篇用于PAT甲级题解,会随着不断刷题持续更新中,至于更新速度呢,嘿嘿,无法估计,不知道什么时候刷完这100多道题. 带*的是我认为比较不错的题目,其它的难点也顶多是细节处理的问题~ 做着做着,发现有些题目真的是太水了,都不想写题解了…

浏览全部代码:请戳 本文谨代表个人思路,欢迎讨论;) 1041. Be Unique (20) 题意 给出 N (<=105)个数(数值范围为 [1, 104]),找到其中不重复的第一个数字.比如给出5 31 5 88 67 88 17 , 答案是 31 . 分析 简单模拟题,开一个大数组int a[10001];,以读入的数为下标,记录 count:a[index] ++;.结果输出第一个存储为 1 的下标:if (a[index] == 1). 1042. Shuffling Machine…

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准备每天刷两题PAT真题.(一句话题解) 1001 A+B Format  模拟输出,注意格式 #include <cstdio> #include <cstring> #include <iostream> #include <algorithm> using namespace std; string ans = ""; int main() { ; cin >> a >> b; c = a + b; ) {…