#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…

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/6789189.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~~ 水题,找出哪个数只出现过一次,输出那个数如果没有的话,输出None #include <iostream> #include <cstdio> #include <algorithm> #include <string.h&g…

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/6789775.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~~ 水题,就是统计n个数的数位和有多少个不同的,并且输出即可. #include <iostream> #include <cstdio> #include <algorithm> #include <string> #in…

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,…

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…

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);…

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…

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);…